1. We’ll show you how to calculate the pH of a solution formed by mixing a strong acid with a strong base. Strong Acid– Strong Base Mixtures

2. We’ll consider adding equal amounts of a strong acid and a strong base to the beaker on the bottom. Strong Acid Strong Base

3. Strong acids contribute hydronium ions to aqueous solutions. H3O+H3O+ Strong Acid Strong Base

4. Acids can more simply be viewed as donating protons or H+ ions. HERE, we’ll consider acids as a source of H+ ions. H+H+ Strong Acid Strong Base

5. And bases are a source of hydroxide or OH minus ions. H+H+ Strong Acid Strong Base OH –

6. When a strong acid is added to a container (click), it brings H+ ions with it. Strong Acid Strong Base OH – H+H+

7. And when a strong base is added (click), it brings hydroxide or OH minus ions with it. Of course when the base is added it immediately mixes with the acid. Strong Acid Strong Base H+H+ OH –

8. However, here we’ll pretend it doesn’t mix yet, so we can get a more detailed idea of the processes. OH – Strong Acid Strong Base H+H+

9. As the solutions mix, H+ and OH minus ions move toward each other. OH – Strong Acid Strong Base H+H+

10. And in the process of neutralization, react with each other to form water. OH – Strong Acid Strong Base H+H+ H2OH2O

11. And the spectator ions in the acid and base will be present as a neutral salt. Strong Acid Strong Base H 2 O and a Salt

12. Now we’ll use a slightly different model to visualize adding a strong acid to a strong base. In this case we’ll start with equal moles (click) of H+ and OH minus, and move these together. Strong Acid H+H+H+H+H+H+H+H+H+H+H+H+ Strong Base OH –

13. The H+ and OH- will neutralize each other to form water. Strong Acid H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ Strong Base OH – H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O

14. and there is no strong acid or base left over, H2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2O

15. so the final mixture is neutral H2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2O The Final Mixture is Neutral

16. This time the acid we’re starting with has more H+ than the base has OH minus. The H+ is in excess.(click) Strong Acid H+H+H+H+H+H+H+H+H+H+H+H+H+H+ Strong Base OH –

17. Each OH- ion will neutralize an H+ ion to form water Strong Acid H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ Strong Base OH – H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O

18. But since we had started with an excess of H+, we have some H+ left over. There weren’t enough OH minus ions available to neutralize all the H+ ions we added. H+H+ H2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2O Excess H + Left Over

19. Because there is H+ left over, the final mixture is Acidic H+H+ H2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2OH2O The Final Mixture is Acidic Excess H + Left Over

20. Now we’ll look at another combination.This time the base we’re starting with has More OH minus than the acid has H+. The OH minus is in excess (click). So we combine them. Strong Base OH – Strong Acid H+H+H+H+H+H+H+H+H+H+

21. Each H+ ion present will neutralize one OH- ion to form water. Strong Base OH – Strong Acid H+H+ H+H+ H+H+ H+H+ H+H+ H2OH2OH2OH2OH2OH2OH2OH2OH2OH2O

22. This time, we have excess OH minus left over. OH – H2OH2OH2OH2OH2OH2OH2OH2OH2OH2O Excess OH – Left Over

23. so the final mixture is basic. OH – H2OH2OH2OH2OH2OH2OH2OH2OH2OH2O Excess OH – Left Over The Final Mixture is Basic

24. Now we’ll do an example question. We’re told that 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH.

25. And we’re asked to find the pH of the final mixture. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

26. We’ll begin by finding the initial moles of H+ added. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

27. Because the source of H+, HCl is a strong monoprotic acid, the moles of H+ is equal to the moles of HCl. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

28. The moles of HCl is equal to 0.200 moles per L 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

29. times 0.0180 L 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

30. Which is 0.00360 moles. So 0.00360 moles of H+ was added. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

31. Now we’ll calculate the initial moles of OH minus added. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

32. Because NaOH is a strong base with one OH in its formula, the moles of OH minus is equal to the moles of NaOH added. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

33. the moles of NaOH is 0.160 moles per L 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

34. times 0.0250 L 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

35. which is 0.00400 moles. So the initial moles of OH minus added is 0.00400 mol. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

36. Notice that both the moles of H+ and OH minus are expressed to 3 significant figures, which is consistent with the given data 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

37. Also notice that both of these values have 5 decimal places. Next we check whether moles of H+ or moles of OH minus is in excess 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

38. 0.00400 moles is greater than 0.00360 moles, so OH minus is in excess. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture. OH – is in Excess

39. Now, 0.00360 moles of H+ will neutralize 0.0036 moles of OH minus. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture. Will neutralize 0.00360 mol OH –

40. So the excess moles of OH minus left over will be 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture. Will neutralize 0.00360 mol OH –

41. The initial 0.00400 moles 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

42. minus the 0.00360 moles of H+ that was able to neutralize 0.00360 moles of OH minus. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

43. Which is equal to 0.00040 moles. And because OH minus is the ion in excess, this is 0.00040 moles of OH minus. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

44. Because the numbers that were subtracted to get this both have 5 decimal places, the value 0.00040 must also have 5 decimal places. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture. 5 decimal places

45. But notice this value, expressed to 5 decimal places, has only 2 significant figures. Zero’s to the left of the first non-zero digit are not significant. So our final answer cannot have more than 2 significant figures. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture. 2 significant figures

46. Because OH minus is the ion in excess, we now calculate the concentration of OH minus in the final mixture. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

47. Concentration of OH minus is moles of OH minus over the total volume in Litres. We have 0.00040 moles of OH minus 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

48. And the total volume of the mixture is 0.0180 L of the HCl solution 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

49. plus 0.0250 L of NaOH solution. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

50. Or 0.0430 L. So the final concentration of OH minus is 0.00040 moles divided by 0.0430 L, 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

51. Which is 0.0093 M. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

52. We want to find the pH of the mixture, but since we have the OH minus concentration, we start by calculating the pOH, which is the negative log of the hydroxide ion concentration 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

53. Or the negative log of 0.0093 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

54. Which is 2.03. Take a moment to check this on your calculator. Remember, 2 decimal places in a pOH value is 2 significant figures. 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

55. In the last step, we’ll calculate the pH 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

56. Which is 14 minus the pOH 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

57. Or 14 minus the 2.03 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

58. Which is 11.97. pH of the final mixture 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.

59. So we’ve now answered what we set out to answer. The pH of the final mixture is 11.97. This is expressed to 2 decimal places, which in a pH value, is 2 significant figures. pH of the final mixture 18.0 mL of 0.200 M HCl is added to 25.0 mL of 0.160 M NaOH. Find the pH of the final mixture.