1. Precipitation Titration Calculations Video Example Here we’ll be given some data from a titration and asked to use this data to calculate the concentration of an ion in a sample.

2. nA nA We’ll start by looking at how we handle titration calculations in general. Titration calculations in Chem 12 involve the reaction between two reactants, which we’ll call A and B here. nBnB mAmA c A & V A c B (given V B ) V B (given c B) mBmB n  moles c  molar concentration (mol/L) m  mass (g) V  volume (L)

3. nA nA In the center of all titration calculations are the moles of reactant A and the moles of reactant B. We represent number of moles in chemistry by the letter n. nBnB mAmA c A & V A c B (given V B ) V B (given c B) mBmB n  moles c  molar concentration (mol/L) m  mass (g) V  volume (L)

4. nA nA Reactant A represents the reactant that we are given enough information to find the number of moles of. nBnB mAmA c A & V A c B (given V B ) V B (given c B) mBmB n  moles c  molar concentration (mol/L) m  mass (g) V  volume (L) Given enough information to find

5. nA nA To convert from moles of A to moles of B, (click) we Always use the mole ratio, or coefficient ratio of B to A in the balanced equation for the titration reaction. nBnB mAmA c A & V A c B (given V B ) V B (given c B) mBmB Mole ratio: B to A n  moles c  molar concentration (mol/L) m  mass (g) V  volume (L)

6. nA nA The information we are given about A could be the molar concentration of A ( represented by the letter c) nBnB mAmA c A & V A c B (given V B ) V B (given c B) mBmB Mole ratio: B to A n  moles c  molar concentration (mol/L) V  volume (L) m  mass (g)

7. nA nA and the volume of A in Litres, represented by the letter V nBnB mAmA c A & V A c B (given V B ) V B (given c B) mBmB Mole ratio: B to A n  moles c  molar concentration (mol/L) V  volume (L) m  mass (g)

8. nA nA or it could be the mass of A in grams, represented by the letter m. nBnB mAmA c A & V A c B (given V B ) V B (given c B) mBmB Mole ratio: B to A n  moles c  molar concentration (mol/L) V  volume (L) m  mass (g)

9. nA nA Whatever we’re given, step 1 of a titration calculation is to convert what we’re given to moles of reactant A, OR N (A) nBnB mAmA c A & V A c B (given V B ) V B (given c B) mBmB Mole ratio: B to A n  moles c  molar concentration (mol/L) V  volume (L) m  mass (g) 1 1

10. nA nA Step 2 of any titration calculation is to convert moles of reactant A to moles of reactant B nBnB mAmA c A & V A c B (given V B ) V B (given c B) mBmB Mole ratio: B to A n  moles c  molar concentration (mol/L) V  volume (L) m  mass (g) 2

11. nA nA This is done using the mole ratio or coefficient ratio of B to A in the balanced equation. nBnB mAmA c A & V A c B (given V B ) V B (given c B) mBmB Mole ratio: B to A n  moles c  molar concentration (mol/L) V  volume (L) m  mass (g) 2

12. nA nA We could be asked one of 3 different things for reactant B. We could be given the volume of B and asked to find its molar concentration c (B). nBnB mAmA c A & V A c B (given V B ) V B (given c B) mBmB Mole ratio: B to A n  moles c  molar concentration (mol/L) V  volume (L) m  mass (g) ?

13. nA nA We could be given the concentration of B and asked to find its volume V (B). nBnB mAmA c A & V A c B (given V B ) V B (given c B) mBmB Mole ratio: B to A n  moles c  molar concentration (mol/L) V  volume (L) m  mass (g) ?

14. nA nA Or we could be asked to find the mass of B in grams. nBnB mAmA c A & V A c B (given V B ) V B (given c B) mBmB Mole ratio: B to A n  moles c  molar concentration (mol/L) V  volume (L) m  mass (g) ?

15. nA nA Step 3 in any titration calculation is to convert moles of B to whatever we’re asked for: concentration of B, volume of B, or mass of B. nBnB mAmA c A & V A c B (given V B ) V B (given c B) mBmB Mole ratio: B to A n  moles c  molar concentration (mol/L) V  volume (L) m  mass (g) 3 3

16. nA nA So here is a generic diagram that outlines the possible steps to take in most titration calculation problems. Remember the first step is always to find moles of what we can. nBnB mAmA c A & V A c B (given V B ) V B (given c B) mBmB Mole ratio: B to A n  moles c  molar concentration (mol/L) V  volume (L) m  mass (g)

17. Let’s do an example precipitation titration question. Example of a Precipitation Titration Calculation Problem

18. A 50.0 mL sample of a solution known to contain chloride (Cl – ) ions is titrated with 0.100 M AgNO 3 solution. A small amount of sodium chromate is added to the sample as an indicator. 0.100 M AgNO 3 A few drops of Na 2 CrO 4(aq) A 50.0 mL sample of a solution known to contain chloride (Cl – ) ions is titrated with 0.100 M AgNO 3 solution. A small amount of sodium chromate (Na 2 CrO 4 ) indicator is added to the sample. 50.00 mL of a solution containing Cl –

19. Three separate trials are done. 0.100 M AgNO 3 50.00 mL of a solution containing Cl – A few drops of Na 2 CrO 4(aq) A 50.0 mL sample of a solution known to contain chloride (Cl – ) ions is titrated with 0.100 M AgNO 3 solution. A small amount of sodium chromate (Na 2 CrO 4 ) indicator is added to the sample. Three separate trials are done.

20. We’re asked to find the concentration of Cl minus in the original sample. The results are recorded in a table. 0.100 M AgNO 3 50.00 mL of a solution containing Cl – A few drops of Na 2 CrO 4(aq) A 50.0 mL sample of a solution known to contain chloride (Cl – ) ions is titrated with 0.100 M AgNO 3 solution. A small amount of sodium chromate (Na 2 CrO 4 ) indicator is added to the sample. Three separate trials are done. Find the [Cl – ] in the original sample

21. Like this. The first thing we need to do is calculate the volume of AgNO3 solution used in each trial. A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample. [AgNO 3 ]=0.100 M Volume of Cl – solution = 50.0 mL Trial 1Trial 2Trial 3 Initial buret reading (mL)0.954.457.65 Final buret reading (mL)4.467.6510.87 Volume of AgNO 3 used (mL)

22. We do that by subtracting the initial buret reading from the final buret reading. So in Trial 1, it is 4.46 minus 0.95 [AgNO 3 ]=0.100 M Volume of Cl – solution = 50.0 mL Trial 1Trial 2Trial 3 Initial buret reading (mL)0.954.457.65 Final buret reading (mL)4.467.6510.87 Volume of AgNO 3 used (mL) A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample. 4.46 – 0.95 = 3.51

23. Which is 3.51 mL [AgNO 3 ]=0.100 M Volume of Cl – solution = 50.0 mL Trial 1Trial 2Trial 3 Initial buret reading (mL)0.954.457.65 Final buret reading (mL)4.467.6510.87 Volume of AgNO 3 used (mL)3.51 A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample. 4.46 – 0.95 = 3.51

24. For trial 2, the volume is 7.65 minus 4.45 [AgNO 3 ]=0.100 M Volume of Cl – solution = 50.0 mL Trial 1Trial 2Trial 3 Initial buret reading (mL)0.954.457.65 Final buret reading (mL)4.467.6510.87 Volume of AgNO 3 used (mL)3.51 A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample. 7.65 – 4.45 = 3.20

25. Which is 3.20 mL [AgNO 3 ]=0.100 M Volume of Cl – solution = 50.0 mL Trial 1Trial 2Trial 3 Initial buret reading (mL)0.954.457.65 Final buret reading (mL)4.467.6510.87 Volume of AgNO 3 used (mL)3.513.20 A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample. 7.65 – 4.45 = 3.20

26. And in Trial 3, the volume used is 10.87 minus 7.65 [AgNO 3 ]=0.100 M Volume of Cl – solution = 50.0 mL Trial 1Trial 2Trial 3 Initial buret reading (mL)0.954.457.65 Final buret reading (mL)4.467.6510.87 Volume of AgNO 3 used (mL)3.513.20 A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample. 10.87 – 7.65 = 3.22

27. Which is 3.22 mL [AgNO 3 ]=0.100 M Volume of Cl – solution = 50.0 mL Trial 1Trial 2Trial 3 Initial buret reading (mL)0.954.457.65 Final buret reading (mL)4.467.6510.87 Volume of AgNO 3 used (mL)3.513.203.22 A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample. 10.87 – 7.65 = 3.22

28. Taking a look at these three results, we see that the volume used in Trial 1, 3.51 mL, is considerably higher than the 3.20 and 3.22 used in trials 2 and 3, respectively. [AgNO 3 ]=0.100 M Volume of Cl – solution = 50.0 mL Trial 1Trial 2Trial 3 Initial buret reading (mL)0.954.457.65 Final buret reading (mL)4.467.6510.87 Volume of AgNO 3 used (mL)3.513.203.22 A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample. Considerably higher than 3.20 and 3.22

29. For that reason, we just discard the value of 3.51 [AgNO 3 ]=0.100 M Volume of Cl – solution = 50.0 mL Trial 1Trial 2Trial 3 Initial buret reading (mL)0.954.457.65 Final buret reading (mL)4.467.6510.87 Volume of AgNO 3 used (mL)3.513.203.22 A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample. DISCARD

30. We calculate the best average volume of AgNO3, by taking 3.20 plus 3.22 and dividing by 2, which gives us [AgNO 3 ]=0.100 M Volume of Cl – solution = 50.0 mL Trial 1Trial 2Trial 3 Initial buret reading (mL)0.954.457.65 Final buret reading (mL)4.467.6510.87 Volume of AgNO 3 used (mL)3.513.203.22 A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample. Best Average =

31. 3.21 mL [AgNO 3 ]=0.100 M Volume of Cl – solution = 50.0 mL Trial 1Trial 2Trial 3 Initial buret reading (mL)0.954.457.65 Final buret reading (mL)4.467.6510.87 Volume of AgNO 3 used (mL)3.513.203.22 A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample. Best Average =

32. We’ll make a note of the average volume of 3.21 mL up here in the table. [AgNO 3 ]=0.100 M Volume of Cl – solution = 50.0 mL Volume of AgNO 3 used: 3.21 mL Trial 1Trial 2Trial 3 Initial buret reading (mL)0.954.457.65 Final buret reading (mL)4.467.6510.87 Volume of AgNO 3 used (mL)3.513.203.22 A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample. Best Average =

33. We’ll convert the 3.21 mL to 0.00321 L [AgNO 3 ]=0.100 M Volume of Cl – solution = 50.0 mL Volume of AgNO 3 used: 3.21 mL = 0.00321 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample.

34. At this point, let’s dissociate the AgNO3 here [AgNO 3 ]=0.100 M Volume of Cl – solution = 50.0 mL Volume of AgNO 3 used: 3.21 mL = 0.00321 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample.

35. And we get Ag+ and NO3 minus [Ag + NO 3 – ] = 0.100 M Volume of Cl – solution = 50.0 mL Volume of AgNO 3 used: 3.21 mL = 0.00321 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample.

36. And we’ll also dissociate the AgNO3 here… [Ag + NO 3 – ] = 0.100 M Volume of Cl – solution = 50.0 mL Volume of AgNO 3 used: 3.21 mL = 0.00321 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample.

37. Giving us Ag+ and NO3 minus [Ag + NO 3 – ] = 0.100 M Volume of Cl – solution = 50.0 mL Volume of Ag + NO 3 – used: 3.21 mL = 0.00321 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample.

38. The nitrate ion, NO3 minus is a spectator ion. It does not form any precipitates. [Ag + NO 3 – ] = 0.100 M Volume of Cl – solution = 50.0 mL Volume of Ag + NO 3 – used: 3.21 mL = 0.00321 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample.

39. So we’ll just discard it. [Ag + NO 3 – ] = 0.100 M Volume of Cl – solution = 50.0 mL Volume of Ag + NO 3 – used: 3.21 mL = 0.00321 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample.

40. So we can simply say that the concentration of Ag+ is 0.100 molar [Ag + ] = 0.100 M Volume of Cl – solution = 50.0 mL Volume of Ag + used: 3.21 mL = 0.00321 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample.

41. And the volume of Ag+ solution used is [Ag + ] = 0.100 M Volume of Cl – solution = 50.0 mL Volume of Ag + used: 3.21 mL = 0.00321 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample.

42. Equal to 0.00321 L [Ag + ] = 0.100 M Volume of Cl – solution = 50.0 mL Volume of Ag + used: 3.21 mL = 0.00321 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find the [Cl – ] in the original sample.

43. So we have all the information we need up here now. We’ll just rearrange it a bit… A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample. [Ag + ] = 0.100 M Volume of Cl – solution = 50.0 mL Volume of Ag + used: 3.21 mL = 0.00321 L

44. So it looks like this. [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 50.0 mL A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

45. We have the concentration of Ag+ and the volume of Ag+ here. [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 50.0 mL A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

46. And the volume of Cl minus here. [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 50.0 mL A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

47. At this point, we’ll convert the 50.0 mL of Cl minus solution [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 50.0 mL = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

48. To 0.0500 litres [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 50.0 mL = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

49. And leave it like this. [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

50. Here is the generic diagram we came up with for titrations. [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

51. We are given the concentration and volume of Ag+ [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

52. And we’re asked for the concentration of the other reactant, Cl minus [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

53. And we’re given its volume. [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

54. We’ll start with the concentration and volume of Ag+ [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

55. Convert that to moles of Ag+ [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

56. Then find moles of Cl minus [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

57. And use that and the given volume, to calculate the concentration of Cl minus [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

58. We can do the first 2 steps using conversion factors. [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

59. We take the concentration of Ag+, which is 0.100 mol Ag+ per 1 litre Ag+ [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

60. and multiply it by the volume of Ag+ used, which is 0.00321 L of Ag+ [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

61. If we cancel out the unit, litres of Ag+ [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

62. we’re left with moles of Ag+ [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

63. so evaluating this would give us moles of Ag+, or n Ag+ [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

64. However, rather than stopping here, we’ll add another conversion factor to get the moles of Cl minus, or n Cl minus [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

65. Looking at the coefficients of Ag+ and Cl minus in the balanced net ionic equation, the mole ratio is 1 mole of Cl minus to 1 mole of Ag+. [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

66. We’ll cancel the unit moles of Ag+ [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

67. And we’re left with the unit moles of Cl minus [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

68. So evaluating this expression will give us the moles of Cl minus, or n Cl minus [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

69. 0.100 times 0.00321 times 1 over 1 gives us 0.000321 moles of Cl minus. [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

70. To convert moles of Cl minus to concentration of Cl minus, we’ll use the formula c equals n over V [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

71. n Cl minus is 0.000321 moles of Cl minus [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

72. And the volume of the Cl minus solution, V Cl minus, is 0.0500 L [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

73. So the concentration of Cl minus in the original sample, c Cl minus, is [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

74. 0.000321 mol of Cl minus over 0.0500 L of Cl minus [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

75. which equals 0.00642 moles per litre. [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

76. or 0.00642 molar [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

77. You may want to pause the video at this point, so you can take a screen shot and use this as a copy of the whole solution to this problem. [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample.

78. So we can summarize by stating that the concentration of Cl minus in the original chloride solution was 0.00642 molar. [Ag + ] = 0.100 M Volume of Ag + used = 0.00321 L Volume of Cl – solution = 0.0500 L A 50.0 mL sample of a solution containing Cl – ions is titrated with 0.100 M AgNO 3. The net ionic equation for the precipitation reaction is: Find [Cl – ] in the original sample. The [Cl – ] in the original sample was 0.00642 M 50.00 mL of a solution containing Cl –