Question

A beaker with 1.90×102 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 7.70 mL of a 0.430 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

let x = concentration acid
let y = concentration conjugate base

x + y = 0.100
5.00 = 4.740 + log y/x

we must solve this system

5.00 - 4.740 = log y/x
0.26 = log y/x
10^0.26 =1.82 = y/x

1.82 x = y

x + 1.82x = 0.100
2.82 x = 0.100
x =0.03546 M = concentration acid
0.100 - 0.03546 = 0.06454 M= concentration conjugate base

Volume = 1.90×102 mL = 193.8 mL = 0.1938 L

moles acid = 0.1938 L x 0.03546 M= 0.006872148
moles conjugate base = 0.06454 M x 0.1938 L =0.012507852

moles HCl = 7.70 x 10^-3 L x 0.430 M=0.003311

A- + H+ = HA

moles conjugate base =0.012507852 - 0.003311 =0.009196852
moles acid =0.006872148+ 0.003311 =0.010183148

total volume = 193.80 + 7.70 = 201.5 mL =0.2015  L

concentration acid = 0.010183148/ 0.2015 =0.0505 M
concentration conjugate base =0.009196852/ 0.2015 =0.0456 M

pH = 4.740 + log 0.0456 / 0.0505 =4.695

change pH = 5.00 -4.695=0.305

Hence , the pH will change by 0.305