Question

A beaker with 135 mLmL 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 MM. A student adds 6.20 mLmL of a 0.480 MM HClHCl solution to the beaker. How much will the pH change? The pKapKa of acetic acid is 4.740.

Answer 1

This problem is divided into categories

(a) calculation of [CH3COOH] AND [CH3COO-]

Handerson Hasselbalch equation will be used here, which is

pH = pKa + log([conjugate base] /[Acid])

for acetic acid buffer

pH =pKa +log([CH3COO-] /[CH3COOH]) ...... (1)

given pH =5.000

pKa =4.740

put in above equation

5.000 = 4.740 + log([CH3COO-] /[CH3COOH])

log([CH3COO-] /[CH3COOH]) = 0.260

now from here

[CH3COO-]/[CH3COOH] = 10^(0.260) = 1.8197

from here [CH3COO-] = 1.8197 * [CH3COOH]. ..... (2)

now given total concentration = [CH3COO-] +[CH3COOH] =0.100

Let [CH3COO-] = x

then from above equation

[CH3COOH] =0.100 - x

put these values in (1)

x = 1.8197 *(0.100-x)

x = 0.18197 - 1.8197*x

2.8197 *x =0.18197

then x = 0.18197 / 2.8197 = 0.064535

thus [CH3COO-] =0.064535

[CH3COOH] =0.100 - x = 0.100 - 0.064535= 0.035465

(b) calculation of moles of CH3COOH AND CH3COO-

given volume of CH3COOH = 135mL =0.135L

moles of CH3COOH =[CH3COOH] * volume =

=0.035465* 0.135 = 0.0047877

moles of CH3COO- = [CH3COO-] *volume =

0.06454 *0.135 = 0.00871226

Thus buffer contains more conjugate base.

(c) to calculate no. of moles of HCl-

volume of HCl added = 6.20mL = 0.0062L

molarity of HCl = 0.480M

no.of moles of HCl added = molarity* volume = 0.0062*0.480 =0.002976

(d) to find change in moles of CH3COOH AND CH3COO- after addition of HCl-

as H3O+ is completely consumed. so

no. of moles of H3O+ =0

as HCl is added, so moles of CH3COO-decreases and moles of CH3COOH increases

after addition

moles of CH3COOH=0.0047877 +0.002976 = 0.0077637

moles of CH3COO- = 0.00871226- 0.002976 = 0.00573626

total volume = 6.20 +135 = 141.20mL =0.1412 L

(e) to calculate [CH3COOH] AND [CH3COO-] after addition of HCl-

[CH3COOH] =moles of CH3COOH /volume

= 0.0077637/0.1412 =0.054979

[CH3COO-] =moles of CH3COO- /volume

=0.00573626/0.1412 =0.040625

(f) finally to calculate pH after addition of HCl and to find change in pH-

now using equation (1)

pH = 4.740 + log (0.040625/0.054979) =4.609

starting pH was =5. 000

so change in pH = 5.000-4.609 = 0.391

comment if any issue, hope u like it.