Question

A beaker with 2.00×10² 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 5.40 mL of a 0.490 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

volume of buffer = 2.00x10^2 mL = 200.0mL =0.200L

total concentration of solution = 0.100M

let be the concentration of the acid = x M

concentration of conjugate base = (0.100 - x)M

number of moles of acid = xM * 0.200L= 0.2x moles

number of moles of conjugate base = (0.100-x)M * 0.200L =( 0.02 - 0.2x)moles

Pka = 4.740

PH= 5.00

PH= PKa +log[conjugate base]/[acid]

5.00 = 4.740 + log(0.02-0.2x)/0.2x

0.26 = log(0.02-0.2x)/0.2x

(0.02-0.2x)/0.2x = 10^0.26

(0.02-0.2x)/0.2x= 1.820

0.02 - 0.2x= 0.364 x

0.02 = 0.564x

x= 0.0355M

Hence

Concentration of acid = 0.0355M

concentration of conjugate base = 0.1-0.0355 = 0.0645M

number of moles of acid = 0.0355M x 0.200L= 0.0071 moles

number of moles of conjugate base = 0.0645M x 0.200L= 0.0129 moles

HCl =5.40 mL of 0.490M

number of moles of HCl = 0.490M x 0.00540L=0.002646 moles

after addition of HCl

number of moles of acid = 0.0071+0.002646=0.009746 moles

number of moles of conjugate base = 0.0129 - 0.002646 = 0.010254 moles

PH = 4.740 +log(0.010254/0.009746)

PH= 4.762

PH= 4.76

Change in PH = 5.00 - 4.76 = 0.24.