Question

A buffer is made by dissolving 4.800 g of sodium formate (NaCHO2) in 100.00 mL of a 0.30 M solution of formic acid (HCHO2). The Ka of formic acid is 1.8 x 10-4.
a) What is the pH of this buffer?
b) Write two chemical equations showing how this buffer neutralizes added acid and added base.
c) What mass of solid NaOH can be added to the solution before the pH rises above 4.60?

Answer 1

a) The moles of base and acid are calculated:

n base = m / MM = 4.8 / 68 = 0.07 mol

n acid = M * V = 0.3 M * 0.1 L = 0.03 mol

The pH is calculated:

pH = - log Ka + log (n base / n acid) = - log 1.8E-4 + log (0.07 / 0.03) = 4.11

b) The reactions are:

CHO2- + HCl = HCHO2 + Cl-

HCHO2 + NaOH = NaCHO2 + H2O

c) Calculate the pKa and the molar ratio of the components:

pKa = - log 1.8E-4 = 3.74

n base / n acid = 10 ^ (pH - pKa) = 10 ^ (4.60 - 3.74) = 7.2

It has:

i) n base - 7.2 * n acid = 0

ii) n base + n acid = 0.03 + 0.07 = 0.1

The system of equations is applied and we have:

n base = 0.088 mol

n acid = 0.012 mol

The required moles of NaOH are calculated:

n NaOH = 0.088 - 0.07 = 0.018 mol

m NaOH = n * MM = 0.018 mol * 40 g / mol = 0.72 g

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