Question

Problem 17.15 Use information from Appendix D in the textbook to calculate the pH of the following solutions.

Part A a solution that is 6.5

Answer 1

1) It is a buffer solution and it's pH is given by henderson-hasselbalch equation.

pKa of propionic acid = 4.87

molarity of propionic acid = 0.08M

molarity of potassium propionate = 0.065M

pH = pKa + log([salt]/[acid])

pH = 4.87 + log(0.065/0.08) = 4.78

2) it is a basic buffer. hence it is easier to find pOH and convert to pH

pKb of trimethylamine = 4.19

pOH = pKb + log([salt]/[base])

pOH = 4.19 + log(0.11/0.08) = 4.33

therefore pH = 14 - 4.33 = 9.67

3) we have an acidic buffer

pKa of acetic acid = 4.74

here total volume = 50 + 50 = 100ml

M₁V₁=M₂V₂

M₁ * 100 = 0.16 * 50

therefore molarity of acetic acid = 0.08 M

similarly for molarity of sodium acetate = 0.2 * 50 /100 = 0.1

now pH = 4.74 + log(0.1/0.08) = 4.84