Question

A beaker with 1.40×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.40 mL of a 0.420 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

pH= PKa+ log [A-]/[HA]

5= 4.74 + log [A-]/[HA]

Where A- [CH3COO- ] is concentration of conjugate buffer and [HA]= concentration of acetic acid

[A-]/[HA] =10 ^0.26=1.82

[A-] = 1.82*[HA] (1)

But is is given that

Given [A}+ [HA] = 0.1     

1.82[HA] + [HA] = 0.1 [HA] (1.82+1)= 0.1 [HA] =0.1/2.82= 0.035461 M

[A-] = 1.82*0.035461= 0.064539 M

Moles of acetic acid in 140 ml =0.035461*140/1000 =0.004965 moles

Moles of acetate ion in 140ml =0.064539*140/1000 = 0.009035 moles

When HCl is added, the following reaction takes place ( HCl completely ionizes)

CH3COO- + H----à CH3COOH

This adds to Acetic acid moles and reduction in moles of acetate

Volume of HCl= 8.4 Molarity =0.42 M

Moles of HCl added = 0.42*8.4/1000 =0.003528 moles

Moles of Acetic acid after addition of HCl =initial moles (0.004965)+ additional moles (0.003528)= 0.008493

Moles of acetate ion after addition of HCl =0.009035 -0.003528 =0.005507

Volume after addition =140+8.4= 148.4 ml=0.1484 L

Molarity of CH3COOH [HA ]= 0.008493/0.1484 L = 0.05723 M

Molarity of [CH3COO-][A-]=0.005507/0.1484= 0.037112

pH= 4.74+ log [ 0.03712/0.05723)=4.74-0.18811=4.55189