Question

1. Calculate the pH at the equivalence point when a solution of 0.10M acetic acid is titrated with a solution of 0.10 M NaOH solution. Given Ka for acetic acid is 1.8 x 10-5 .

2. The pH of blood serum is maintained by a proper balance of H2CO3 and NaHCO3 concentrations. Calculate the volume of 5M NaHCO3 solution that should be mixed with a 10 mL of sample of blood which is 2M in H2CO3 in order to maintain a pH of 7.4 . Given Ka1 and Ka2 for H2CO3 in blood are 4.3 x 10-7 and 5.6 x 10-11

Answer 1

solution:

Q1:

Acetic acid reacts with NaOH as

CH₃COOH + NaOH <-------> CH₃COO^- + Na⁺ + H₂O

At equivalent point all the )VmL of 0.1L CH₃COOH is neutralised by VmL of 0.1M NaOH to form CH₃COO^- , Na+ and H2O. CH₃COO^- is basic and reacts with water as

CH₃COO^- + H2O <--------------> CH₃COOH + OH^-

The ICE equation is shown below

The concentration of CH₃COO^- will be half of CH₃COOH. Because total volume is double now

Total volume = volume of acid (V mL) + volume of Base (VmL) = 2 VmL

Hence initial concentration of CH₃COO^- = 0.1 / 2 = 0.05M

CH₃COO^- <--------------> CH₃COOH + OH^-  

I 0.05 0 0

C -x +x +x

E 0.05 - x x x

K_b = [CH₃COOH] [OH^-] / [CH3COO^-]

K_b = [x] [x] / [0.05 - x]

Kb X Ka = Kw

Kb = Kw / Ka = 10^-14 / 1.8 X 10^-5 = 5.5 X 10^-10

5.5 X 10^-10 =  [x] [x] / [0.05 - x]

Since LHS is very small, x can be neglected in the denominator of RHS

x² = 5.5 X 10^-10 X 0.05 = 2.75 X 10^-11

x = 5.24 X 10^-6

Therefore [OH-] = 5.24 X 10^-6

[H+][OH-] = 10^-14

[H+] = 1.9 X 10^-9

pH = -log[H+] = -log[1.9 X 10^-9] = 8.72

Plz send the other one separately.