Question

A beaker with 145 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 mol L−1. A student adds 4.90 mL of a 0.340 mol L−1 HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

Answer 1

let a = concentration acid
let b = concentration conjugate base

We know, The total molarity of acid and conjugate base in this buffer is 0.100 mol L−1.

a + b = 0.100

buffer pH = 5

The pKa of acetic acid is 4.760

5.00 = 4.760 + log b/a

5.00 - 4.760 = log b/a

0.24 = log b/a

1.74 = b/a

1.74 a = b

a + 1.74 a = 0.100

2.74 a = 0.100

a = 0.0365 M [concentration acid]

0.100 - 0.0365 =0.0635 M [concentration conjugate base]

moles acid = 0.145 L x 0.0365 M = 5.2925*10^-3
moles conjugate base = 0.0635 M x 0.145 L=9.2075*10^-3
moles HCl = 4.90 x 10*^-3 L x 0.340 M=1.666*10^-3
A^- + H⁺ = HA
moles conjugate base = 9.2075*10^-3 - 1.666*10^-3 = 7.5415*10^-3

moles acid = 5.2925*10^-3 + 1.666*10^-3 = 6.9585*10^-3

total volume = 145 + 4.90 = 149.9 mL = 0.1499 L

concentration acid [a]= 6.9585*10^-3 / 0.1499 =0.04642 M

concentration conjugate base [b]= 7.5415*10^-3/ 0.1499 =0.0503 M

pH = 4.760 + log (0.0503/ 0.04642) =4.7969

change in pH = 5.00 - 4.7969=0.2031

change in pH is 0.2031