Question

A beaker with 1.10×10^2 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 7.00 mL of a 0.490 mol L−1 HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

Please show where all numbers comes from.

Answer 1

pH of buffer solution is calculated by using Henderson's equation pH = pKa + log [ conjugate base ] / [ acid ]

pH = pKa + log [ CH₃COO ^- ] / [ CH₃COOH ]

Substituting pH = 5.000 and pKa = 4.760 , we get

5.000 = 4.760 + log [ CH₃COO ^- ] / [ CH₃COOH ]

log [ CH₃COO ^- ] / [ CH₃COOH ] = 5.000 - 4.760

log [ CH₃COO ^- ] / [ CH₃COOH ] = 0.240

Taking anti log on both sides, we get

[ CH₃COO ^- ] / [ CH₃COOH ] = 10 ^0.240 = 1.74

[ CH₃COO ^- ] = 1.74 [ CH₃COOH ]

We have given , [ CH₃COO ^- ] + [ CH₃COOH ] = 0.100 M ( 1)

Substituting [ CH₃COO ^- ] = 1.74 [ CH₃COOH ] in equation 1 , we get

1.74 [ CH₃COOH ] +  [ CH₃COOH ] = 0.100 M

2.74   [ CH₃COOH ] = 0.100 M

[ CH₃COOH ] = 0.100 M / 2.74

[ CH₃COOH ] = 0.0365 M

We have, [ CH₃COO ^- ] + [ CH₃COOH ] = 0.100 M

[ CH₃COO ^- ] +0.0365 M = 0.100 M

[ CH₃COO ^- ] = 0.100 M - 0.0365 M = 0.0635 M

[ CH₃COO ^- ] = 0.0635 M , [ CH₃COOH ] = 0.0365 M

Now, calculate moles of Acetate ions and acetic acid present in buffer solution.

We know that,[ CH₃COO ^- ] = No. of moles of CH₃COO ^- / volume of solution in L

No. of moles of CH₃COO ^- = [ CH₃COO ^- ] volume of solution in L

No. of moles of CH₃COO ^- = 0.0635 mol / L 0.110 L = 0.006985 mol

Similarly, No. of moles of CH₃COOH = 0.0365 mol / L 0.110 L = 0.004015 mol

Moles of HCl added to buffer solution = 0.490 mol / L 0.00700 L = 0.00343 mol

Consider reaction of HCl with buffer solution.

CH₃COO ^- + HCl   CH₃COOH + Cl -

Let's use ICE table.

moles	CH3COO - + HCl   CH3COOH
I	0.006985	0.00343	0.004015
C	-0.00343	-0.00343	+0.00343
E	0.003555	0.00000	0.007445

Volume of solution after addition of HCl = 110 ml +7.00 ml =117 ml

pH of buffer solution after addition of HCl = 4.760 + log ( 0.003555 mol / 0.117 L ) / ( 0.007445 mol / 0.117 L )

pH = 4.760 - 0.321

pH = 4.439

ANSWER : pH of buffer solution after addition of HCl = 4.439