Question

calculate the pH of the buffer that results from mixing 51.5 mL of a 0.398 M solution of HCHO2 and 11.5 mL of a 0.605 M solution of NaCHO2. the Ka value for HCHO2 is 1.8 x 10^-4

Answer 1

Buffer solution is a solution which resists the change in pH.

Buffer solutions are of two types.

1. Acidic buffer: It is a mixture of weak acid and salt formed by this weak acid with a strong base.

Ex: Acetic acid and sodium acetate (salt of acetic acid with NaOH )

2. Basic buffer: It is a mixture of weak base and salt formed by this weak base with a strong acid.

Ex : Ammonium hydroxide and Ammonium chloride

The given buffer is an acidic buffer ( formic acid and sodium formate)

The pH of buffer is given by Henderson's equation.

Or

Molarity is the number of moles of solute present in one litre of the solution.

M = Number of moles of solute * 1000/Volume of the solution

Number of moles of solute = M * Volume of solution in mL /1000

Number of moles of formic acid in 51.5 mL of 0.398 M formic acid solution is

Number of moles of formic acid =

0.398 * 51.5/1000

= 0.0205

Number of moles of sodium formate in 11.5 mL of 0.605 M sodium formate is

Number of moles of sodium formate =

0.605 * 11.5/1000

= 0.00696

Given that Ka of formic acid = 1.8×10-4

pKa = -logKa = -log(1.8×10^(-4))

= 4 -log1.8

= 4 - 0.255 = 3.745

pH of the given buffer is

= 3.745 + log(0.339)

= 3.745 - 0.469 = 3.276