1. Example Dilution Problem with pH Here’s a problem that deals with diluting a solution to get a desired pH.

2. A 250.0 mL sample of HBr has a pH of –0.40. We’re given that A 250.0 mL sample of HBr has a pH of –0.40.

3. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? And we’re asked how much water needs to be added to the sample to bring the pH up to +0.40?

4. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? Looking at HBr, we see that it’s a Strong Acid, so the [HBr] = [H 3 O + ]. So any [H 3 O + ] in the solution to this problem will be equivalent to the [HBr] HBr is a Strong Acid So [HBr] = [H 3 O + ]

5. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? The best way to organize our information is with a table like this. pH[H 3 O + ] Volume (mL) Initial–0.40V 1 = 250.0 mL Final+0.40V 2 = ?

6. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? We are given the initial and final pH values pH[H 3 O + ] Volume (mL) Initial–0.40V 1 = 250.0 mL Final+0.40V 2 = ?

7. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? But dilution calculations require molar concentrations rather than pH, so pH’s must be converted to hydronium ion concentrations. pH[H 3 O + ] Volume (mL) Initial–0.40V 1 = 250.0 mL Final+0.40V 2 = ?

8. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? Because a dilution calculation will be done, we also need to work with initial and final volumes. pH[H 3 O + ] Volume (mL) Initial–0.40V 1 = 250.0 mL Final+0.40V 2 = ?

9. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? The initial pH is -0.40 pH[H 3 O + ] Volume (mL) Initial–0.40V 1 = 250.0 mL Final+0.40V 2 = ?

10. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? And the final pH is +0.40 pH[H 3 O + ] Volume (mL) Initial–0.40V 1 = 250.0 mL Final+0.40V 2 = ?

11. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? The initial volume, which we’ll call V1 is 250.0 mL. pH[H 3 O + ] Volume (mL) Initial–0.40V 1 = 250.0 mL Final+0.40V 2 = ?

12. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? Given the information here, we can find what the final volume of the solution has to be, in order to have the desired pH of +0.40. We’ll call this the final volume, V2. pH[H 3 O + ] Volume (mL) Initial–0.40V 1 = 250.0 mL Final+0.40V 2 = ? What is the final volume after the required water is added?

13. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? Later, we can calculate how much water needs to be added to reach this volume. pH[H 3 O + ] Volume (mL) Initial–0.40V 1 = 250.0 mL Final+0.40V 2 = ? Later, we can calculate how much water needs to be added to reach this volume.

14. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? In order to do a dilution calculation, we must (click) change both pH’s into hydronium ion concentrations. pH[H 3 O + ] Volume (mL) Initial–0.40V 1 = 250.0 mL Final+0.40V 2 = ?

15. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? We’ll call the initial hydronium ion concentration c 1 pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ]V 1 = 250.0 mL Final+0.40V 2 = ?

16. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? The formula we use is [H 3 O + ] = 10 –pH. pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 –pH V 1 = 250.0 mL Final+0.40V 2 = ?

17. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? Which is 10 to the negative of negative 0.40 pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 –(–0.40) V 1 = 250.0 mL Final+0.40V 2 = ?

18. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? Which is 10 to the positive 0.40 pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 V 1 = 250.0 mL Final+0.40V 2 = ?

19. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? Which comes out to 2.5 M pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40V 2 = ?

20. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? Now we find the hydronium concentration in our final solution, which we’ll call c2 pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ]V 2 = ?

21. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? Again, we use the formula [H 3 O + ] = 10 –pH pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –pH V 2 = ?

22. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? This time, we substitute positive 0.40 in for the pH, so 10 to the negative pH is 10 to the negative 0.40 pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 V 2 = ?

23. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? Which is 0.40 M. 0.40 happens to be the antilog of -0.40. This is a rare case where the antilog of a number is just the negative of the number, when rounded to significant figures. pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = ?

24. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? We now have the values for C1, C2, and V1 pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = ?

25. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? And with a dilution calculation, we can find the value for the final volume, V2. pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = ?

26. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? We’ll start by writing the dilution formula c1V1 = c2V2 or c2V2 = c1V1 pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = ? The Dilution Formula

27. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? Now, we’ll rearrange the equation to solve for our unknown, V2, which is c1V1 over C2 pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = ?

28. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? We can substitute 2.5 M in for C 1 pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = ?

29. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? And 250 mL in for V1 pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = ?

30. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? And finally, we’ll substitute 0.40 M in for C2 pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = ?

31. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? We can now use this expression to calculate the final volume V2. pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = ? Calculate V 2

32. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? We’ll cancel out the unit molarity pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = ? Calculate V 2

33. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? Performing the calculation gives us 1562.5 mL. We won’t round to significant figures until the last step of the problem. pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = ? V2V2

34. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? We’ll record the value for the final volume, V2, in our table here. pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = 1562.5 mL

35. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? When we’re adding water to a solution, the final volume, V2, will be equal to the initial volume, V1 plus the volume of the water added. pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = 1562.5 mL

36. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? We’ll rearrange the equation to solve for the water that needs to be added. This is what we’re asked for in the question. pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = ?

37. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? We’ll substitute 1562.5 mL in for V2 in our equation pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = 1562.5 mL

38. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? And 250.0 mL in for V1 pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = 1562.5 mL

39. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? Subtracting 250.0 mL from 1562.5 mL, gives us 1312.5 mL pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = 1562.5 mL

40. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? The lowest number of significant figures given in our data was 2, so rounding 1312.5 to 2 significant figures, gives us 1300 mL pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = 1562.5 mL

41. A 250.0 mL sample of HBr has a pH of –0.40. How much water needs to be added to the sample to bring the pH up to +0.40? Or 1.3 L pH[H 3 O + ] Volume (mL) Initial–0.40c 1 = [H 3 O + ] = 10 0.40 = 2.5 MV 1 = 250.0 mL Final+0.40c 2 = [H 3 O + ] = 10 –0.40 = 0.40 MV 2 = 1562.5 mL

42. A 250.0 mL sample of HBr has a pH of –0.40. In order to bring the pH up to +0.40, we would need to add 1300 mL, or 1.3 L of water to the original solution. So now we can summarize the final answer. If a sample of the strong acid HBr has a pH of 0.40, we would need to add 1300 mL or 1.3 L of water to this solution to bring the pH up to +0.40