1. Here we’ll go over an example in which a strong acid is mixed with a strong base, and we calculate the pH of the final mixture. Strong Acid–Strong Base Mixture Calculations Example 2

2. We’re given that 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH.

3. And we’re asked to determine the pH of the final mixture. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

4. Just a few words about sulphuric acid, H2SO4. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

5. H2SO4. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? H 2 SO 4

6. is a Diprotic Acid 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? H 2 SO 4 is a Diprotic Acid

7. Which means it has 2 protons it can lose. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? H 2 SO 4 is a Diprotic Acid It has 2 protons it can lose.

8. As soon as H 2 SO 4 is added to water, it ionizes completely to lose its first proton: 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? When H 2 SO 4 is added to water, it ionizes completely to lose its first proton:

9. 100% of the H2SO4 molecules lose one proton (click) to form hydronium and hydrogen sulphate ions. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? When H 2 SO 4 is added to water, it ionizes completely to lose its first proton: H+H+

10. But when its just in water, the second proton does not come off as easily. This proton comes off when HSO4 minus ionizes. But HSO4- is a weak acid so its ionization in water is very limited. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? When H 2 SO 4 is added to water, it ionizes completely to lose its first proton: But its second proton does not come off as easily in water: Equilibrium Weak Acid

11. However, when H 2 SO 4 is mixed with the STRONG BASE KOH, this is a totally different situation. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? When H 2 SO 4 is mixed with the strong base KOH, this is a totally different situation.

12. When an H2SO4 molecule enters water, it loses one proton (click) to water, to form a hydronium ion (H3O+) and a hydrogen sulphate ion (HSO4 minus). 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? H+H+

13. Models of these are shown here. Take a moment to check the atoms and the charges and see how the formulas relate to the structural models. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? O H H H + O O H S O O – H+H+

14. When the strong base KOH dissociates in water it forms K+ and OH minus ions. Here we doubled everything in the equation. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? 2KOH  2K + + 2OH – O H H H + O O H S O O –

15. we show models of the two hydroxide ions from the KOH 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? O H H O – – 2KOH  2K + + 2OH – O H H H + O O H S O O –

16. One of the hydroxide ions collides with the hydronium ion 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? H O O H – – O H H H + O O H S O O –

17. and takes away a proton, to form 2 water molecules 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? H O – O H H H O H O O H S O O –

18. The other hydroxide ion collides with the hydrogen sulphate ion (click) 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? H O H O H – O H H O O H S O O –

19. And takes a proton from it to form a water molecule and a suphate ion 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? O O S O O – H O H H O H – O H H

20. the sulphate ion has the formula SO4 2 minus. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? H O H H O H O O S O O – – Sulphate SO 4 2– O H H

21. So, in an indirect way, 2 hydroxide ions are able to remove both protons from a molecule of H2SO4. We‘ll show this with equations. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? Two OH – ions were able to remove both protons from H 2 SO 4.

22. As soon as H2SO4 is added to water it ionizes completely to form a hydronium ion and a hydrogen sulphate ion. We’ll call this Step 1 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? Two OH – ions were able to remove both protons from H 2 SO 4. Step 1

23. When we add a strong base, one OH minus ion neutralizes the hydronium ion to form 2 water molecules. We’ll call this step 2 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? Two OH – ions were able to remove both protons from H 2 SO 4. + Step 2

24. And the other OH minus ion reacts with hydrogen sulphate to form water and a sulphate ion. We’ll call this Step 3. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? Two OH – ions were able to remove both protons from H 2 SO 4. ++ Step 3

25. Even though we know these 3 steps occur when we add H2SO4 to water and then add a strong base, 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? Even though we know that these steps occur… 1 2 3

26. We can represent the process with a net overall equation: H2SO4 plus 2 OH minus form 2H2O plus SO4 2minus. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? Even though we know that these steps occur… We can represent the process with a net overall equation. 1 2 3

27. so in the overall net reaction, we see that each H2SO4 (click) donates 2 protons or H+ ions to the hydroxide ions. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? 2 H + Each H 2 SO 4 donates 2 protons to the OH– ions

28. From this, we can write the conversion factor stating there are 2 moles of H+ per 1 mole of H2SO4. We can use this conversion factor in any calculation where H2SO4 reacts with a strong base. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? 2 H + Each H 2 SO 4 donates 2 protons to the OH– ions

29. Now we’ll do the calculations for this problem. We’ll begin by calculating the initial moles of H+ added. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

30. Its equal to 0.150 moles of H2SO4 per Litre… 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

31. Times 2 moles of H+ to 1 mole of H2SO4… 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

32. Times 0.125 L 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

33. Which comes to 0.0375 moles of H+. Notice moles of H2SO4 and Litres cancel out. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

34. In order to preserve 3 significant figures (the lowest number of significant figures in the given data)… 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? 3 significant figures

35. The answer to this must be expressed to 4 decimal places. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? 4 decimal places

36. Now we’ll calculate the initial moles of OH minus added. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

37. It is equal to 0.200 moles of KOH per L 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

38. Times 1 mole of OH minus to 1 mole of KOH 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

39. Times 0.150 L 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

40. Which comes out to 0.0300 moles of OH minus. You can see that moles of KOH and Litres both cancel. Notice we also have 3 significant figures and 4 decimal places in this answer. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

41. Now we compare the initial moles of H+ and OH minus. We see that 0.0375, the moles of H+, is greater than 0.0300, the moles of OH minus 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

42. So the H+ is in excess and the OH minus is the limiting reagent. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? In Excess Limiting Reagent

43. We calculate the excess moles of H+… 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

44. By taking 0.0375 moles of H+ 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

45. and subtracting 0.0300 moles of OH minus. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

46. To give us 0.0075 moles of H+ in excess 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

47. This answer, 0.0075, has 4 decimal places, because the numbers we subtracted both had 4 decimal places. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? 4 decimal places

48. But we can see that written this way, this number has only 2 significant figures, the 7 and the 5. Therefore the final answer to this problem cannot have more than 2 significant figures. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? 2 significant figures

49. The next step on the way to pH, is to find the hydronium ion concentration. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

50. Which is equal to the concentration of H+. These are synonymous in chemistry dealing with aqueous solutions. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

51. The concentration of H+ is equal to moles of H+ per Litre of solution. The moles of H+ is 0.0075 moles. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

52. And the total volume of the mixture is 0.125 L of H2SO4 …. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

53. Plus 0.150 L of KOH. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

54. So the concentration of H+ or H3O+ is 0.0075 moles over 0.275 L. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

55. Which comes out to 0.0273 molar. We’ll carry one more significant figure than the 2 our final answer is limited to. We’ll round to 2 significant figures at the end. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

56. In the last step, we’ll find the pH of the mixture. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

57. Remember, pH is defined as the negative log of the hydronium ion concentration. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

58. Which is the negative log of 0.0273 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture?

59. , which comes out to 1.56. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? 2 significant figures

60. In a pH, the digits to the right of the decimal are significant. So this answer has 2 significant figures. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? 2 significant figures

61. Now we have answered the original question. The pH of the final mixture is 1.56. This low value means the solution is fairly acidic. This is reasonable because a strong acid is in excess in this case. 125.0 mL of 0.150 M H 2 SO 4 is mixed with 150.0 mL of 0.200 M KOH. What is the pH of the final mixture? pH of Final Mixture