Question

# A 5.171 g sample of a solid, weak, monoprotic acid is used to make a 100.0...

A 5.171 g sample of a solid, weak, monoprotic acid is used to make a 100.0 mL solution. 26.00 mL of the resulting acid solution is then titrated with 0.09671 M NaOH. The pH after the addition of15.00 mL of the base is 5.51, and the endpoint is reached after the addition of 47.85 mL of the base.

(b) What is the molar mass of the acid?

(c) What is the pKa of the acid?

moles of H+ = moles of OH-
Since acid is monoprotic and the base is monohydroxy,
mole acid = mole base.

Mole base = M x V
n(base) = (0.09671 mol/L)(0.04785 L) = 0.0046mol

Therefore the mole of the acid neutralized will also be 0.0046 mol.
This mole of acid is present in 26 mL of acid solution. Because it was titrated by the base.

Therefore, the mole of acid in 100 mL solution;
0.0046 mol x (100 mL / 26mL) = 0.0177 mol

b) Since 0.0177 mol acid has a mass of 5.171 g, the molar mass of the acid;
5.171 g / 0.0177 mol = 292.15 g/mol

c) The ionization of the weak acid:
HA(aq) ------> H+(aq) + A-(aq)

When 15 mL of NaOH is added, acid is partly neutralized.
n(NaOH) = M x V = (0.09671 mol/L)(0.015 L) = 0.00145 mol.
n(HA) neutralized = 0.00145 mol.
n(HA) unneutralized = 0.0177 mol - 0.00145 mol = 0.01625 mol

Molarity of the remaining acid :
n = 0.01625 mol
V = 26 mL (acid) + 15 mL (base added) = 41 mL

M = n / V = 0.01625 mol / 0.041 L = 0.396 M

pH = 5.51
[H+] = 10^-pH = 10^-5.51 = 3.09x10^-6 M

This means that 0.396 M acid produces 3.09x10^-6 M H+ ion.

HA(aq) --------------------> H+(aq) + A-(aq)
0.396 - 3.09x10^-6 M.... 3.09x10^-6 M.. 3.09x10^-6 M

Ka = [H+][A-] / [HA]
Ka = (3.09x10^-6 )(3.09x10^-6 ) / (0.396)
Ka = 2.41x10^-11

pKa = - log Ka = - log (2.41x10^-11) = 10.62