Question

Determine the pH (to two decimal places) of a solution prepared by adding 59.7 mL of 0.356 M butanoic acid (C3H7COOH) to 0.923 g of sodium butanoate (NaC3H7COO). Assume the volume of the solution does not change and that the 5% approximation is valid. Ka butanoic acid is 4.82

Answer 1

a solution prepared by adding 59.7 mL of 0.356 M butanoic acid (C3H7COOH) to 0.923 g of sodium butanoate (NaC3H7COO).

pKa of butanoic acid = 4.82

Ka = 10 ^-pKa

= 10 ^-4.82

= 1.51*10^-5

Molar mass of sodium butanoate = 110.1 g/mole

Moles of sodium butanoate = 0.923 g / 110.1 g/mole

= 0.008

Concentration of sodium butanoate = 0.008 / 0.0597

= 0.134 M

Butanoic acid dissociates as:

But-H    But- + H+
Initially 0.356 0 0
Finally 0.356 - x 0.134 + x x

Ka = [But-][H+] / [But-H]

1.51*10^-5 = x*(0.134 - x) / (0.356 - x)

1.51*10^-5 = x*(0.134 - x) / 0.356 {since x is very small, so 0.356 - x = 0.356}

x*(0.134 - x) = 0.356 * 1.51*10^-5

x = 4.01*10^-5

[H+] = 4.01*10^-5 M

We know that,

pH = - log [H+]

= - log (4.01*10^-5)

= 4.39

Hence, pH = 4.39