1. Here we’ll go over an example where 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 1

2. We’re given that 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3.

3. And we’re asked to calculate the pH of the final mixture. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

4. We’ll point out something important here. Strontium hydroxide is a strong base, 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

5. and its formula, Sr(OH)2, has 2 OH’s in it. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

6. So when we write the balanced dissociation equation for Sr(OH)2. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture. Dissociation Equation

7. We see that there are 2 moles of OH- for each mole of Sr(OH)2. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture. A conversion factor

8. So we can use this as a conversion factor 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture. A conversion factor

9. We’ll start by calculating the initial moles of OH minus added. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

10. We take 0.200 moles of Sr(OH)2 per L 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

11. Multiply it by the conversion factor 2 mol OH- over 1 mole Sr(OH)2. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

12. And by 0.150 L. We rounded this to 3 significant figures to save room here. The concentration 0.200 mol/L is only 3 significant figures so the answer to this calculation is limited to 3 significant figures. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

13. When we multiply all three numbers we get 0.0600 mol. So the initial moles of OH minus added is 0.0600 mol. Notice, this is expressed to 3 significant figures, which is consistent with the given data. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

14. Our next step is to calculate the initial moles of H+ added. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

15. The H+ comes from the strong acid nitric acid, or HNO3. Each HNO3 releases 1 proton, so we take 0.100 mol HNO3 per L 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

16. Times 1 mol of H+ per 1 mol of HNO3 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

17. Times 0.350 L. Again we rounded this to 3 significant figures to save room. The number of significant figures in the answer is limited by the 3 significant figures in 0.100 M. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

18. The answer comes out to 0.0350 mol. So the initial moles of H+ added is 0.0350 moles. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

19. Notice that in preserving 3 significant figures, both of these have 4 decimal places 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

20. Comparing the initial moles of OH minus with the initial moles of H+, we see that we have more moles of OH minus than of H+, so the OH minus is in excess in this case. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture. Excess

21. The excess moles of OH minus 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

22. Is 0.0600 mol OH minus 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

23. minus 0.0350 mol H+ 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

24. Which equals 0.0250 mol OH minus 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

25. Notice the numbers we’re subtracting both have 4 decimal places, so our answer must also have 4 decimal places. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture. 4 decimal places

26. When expressed to 4 decimal places, this number has 3 significant figures. The zero’s to the left of the 2 are not significant, but the zero after the 5 is. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture. 3 significant figures

27. Since the hydroxide ion is in excess, we calculate its concentration in the final mixture. It is equal to 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

28. 0.0250 moles 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

29. Divided by the total volume of the solution, which is 0.150 L of Strontium hydroxide solution, 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

30. Plus 0.350 L of HNO3 solution. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

31. So the concentration of OH minus is 0.0250 moles over 0.500 L 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

32. Which Equals 0.0500 molar 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

33. Because we have base in excess, we can calculate the pOH 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

34. Which is the negative log of the hydroxide ion concentration 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

35. Or the negative log of 0.0500 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

36. Which comes out to 1.301. None of our data or calculations in this problem have less than 3 significant figures, so the pOH has 3 significant figures or three decimal places. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

37. Now we can calculate the pH 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

38. Which is 14 minus the pOH 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

39. Or 14 minus 1.301 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

40. Which is 12.699. This must has 3 significant figures so its expressed to 3 decimal places. 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.

41. We have now answered the question we set out to answer. The pH of the final mixture is 12.699. This is relatively highly basic which is consistent with the fact that OH- is in excess. pH of final mixture 150.0 mL of 0.200 M Sr(OH) 2 is added to 350.0 mL of 0.100 M HNO 3. Calculate the pH of the final mixture.