4 Calculate the pH of a solution prepared by mixing 3000 mL

4. Calculate the pH of a solution prepared by mixing 30.00 mL of a 0.15-M KOH solution with 50.00 mL of a .39-M HBr solution. Â

5. Â 45.0 mL of a 0.10-M formic acid (HCHO2, Ka=1.7x10^-4) solution was titrated with a 0.20-M NaOH solution. Â What is the equivalent volume? Â Calculate which is in excess.

6. Â What is the pH of the acidic solution before titration?

7. Â What is the pH of the reaction mixture (in Question 6) after 10.0 mL of NaOH was added?

8. Â What is the pH of the reaction mixture (titration in Question 6) at the equivalence point?

9. Â What is the pH of the reaction mixture (titration in Question 6) after 25.0 mL of base was added?

Solution

4) Moles of KOH = 30.00/1000 L x 0.15 mol/L = 0.0045 mol
Moles of HBr = 50.00/1000 L x 0.39 mol/L = 0.0195 mol
0.0045 mol KOH reacts with 0.0045 mol HBr.
Remaining moles of HBr = 0.0195 - 0.0045
= 0.015
Volume of the solution = 30.00 + 50.00 = 80.00 mL
= 0.08 L
[HBr] = 0.015/0.08 = 0.1875
HBr is a strong acod.
Therefore [H+] = [HBr] = 0.1875
pH = -log[H+] = -log(0.1875) = 0.727
Ans: 0.727
(Note: You can get more accurate result by using Ka for HBr. I did not use because it is not given in the problem.)

5) 1 mol HCHO2 reacts with 1 mol NaOH
Therefore M(formic acid) x V(formic acid)
= M(NaOH) x V(NaOH)
0.10 x 45.0 = 0.20 x V(NaOH)
V(NaOH) = 0.10 x 45.0/0.20 = 22.5 mL
Ans: 22.5 mL

6) HCHO2 <------> H+ + CHO2 -
.......0.10..................0........… (initially)
.........-x.....................x.....… (reaction)
......0.10-x................x.........… (equilibrium)

(x)(x)/(0.10-x) = Ka = 1.7 x 10^-4
Formic acid is a weak acid. Therefore x << 0.10
x^2/0.10 = 1.7 x 10^-4
x^2 = 1.7 x 10^-5
[H+] = x = sqrt(1.7 x 10^-5) = 0.0041231
pH = -log[H+] = -log(0.0041231) = 2.38
Ans: 2.38