Question

Determine the mass of C₆H₄(CO₂H)₂ and the volume of 0.400 M NaOH solution required to make 1.10 L of a buffer solution at pH=3.00 and at a total concentration of 0.38 M. pK_a1(C₆H₄(CO₂H)₂)=2.95.

1. Mass of C₆H₄(CO₂H)₂ (g):

2. Volume of NaOH solution (L):

Answer 1

Mass of C6H4(CO2H)2 = 32.750g

Volume of NaOH solution = 0.5522L

Explanation

Henderson - Hasselbalch equation is

pH = pKa + log ([HA-]/[H2A])

3.00= 2.95 + log ( [HA-]/[H2A])

log([HA-]/[H2A] ) = 0.05

[HA-]/[H2A] = 1.12

[ HA- ] = 1.12[H2A]

Total buffer concentration is 0.38M

[HA-] + [ H2A] = 0.38M

1.12[H2A] + [ H2A] = 0.38M

2.12[ H2A] = 0.38M

[ H2A ] = 0.1792M

[ HA- ] = 0.38M - 0.1792M = 0.2008M

So, Concentration requirements are

[ HA- ] = [ C6H4(COO- ) (CO2H) ] = 0.2008M

[ H2A] = [ C6H4(CO2H)2] =0.1792M

The reaction is

C6H4(CO2H)2 + NaOH ------> C6H4(COO-)(CO2H) + H2O

this reaction is 1:1 mol reaction

So, 0.2008mol of NaOH is required to form 0.2008mol of HA- and remaining mole of H2A will be 0.1792mol

for 1.10L buffer

mole of H2A required = (0.1792mol/1000ml)×1100ml = 0.19712mol

mole of NaOH required = (0.2008mol/1000ml) × 1100ml = 0.22088mol

molar mass of C6H4(CO2H)2 =166.14g/mol

mass of C6H4(CO2H)2 required = 166.14g/mol ×0.19712mol =32.750g

Volume of NaOH solution required = (1000ml/0.400mol)×0.22088mol = 552.2ml