Question

A beaker with 195 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 6.20 mL of a 0.410 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.  

Answer 1

Given the volume of acetic acid buffer = 195 mL

Total molarity of acetic acid and sodium acetate = 1.00 M

Let the molarity of acetic acid = x

therefore molarity of sodium acetate = 1-x

Given pH of buffer is 5

pKa of acetic acid = 4.74

Apply Henderson equation,

pH = pKa + log [salt]/[acid]

5 = 4.74 + log [1-x]/[x]

log 1-x /x = 0.26

1-x /x = 10^0.26

   1-x /x = 1.82

2.82 x = 1

x = 1/2.82 = 0.3546

Thus the molarity of acetic acid = 0.3546 M

and the molarity of sodium acetate = 1-0.3546 = 0.6454 M

Therefore number of moles of acetic acid = 195 mL x 0.3546 moles / 1000 mL

                                                           = 0.069147 moles

Similarly number of moles of sodium acetate = 195 mL x 0.6454 moles / 1000 mL

                                                                = 0.125853 moles

Given that a student adds 6.2 mL of 0.410 M HCl solution.

6.2 mL of 0.410 M HCl = 6.2 x 0.410 / 1000 moles of HCl

                                 = 0.002542 moles

consider the reaction

CH₃COONa + HCl CH₃COOH + NaCl

Thus 1 mole of HCl convert 1 mole of salt in to acid. Therefore after the addition ofHCl, the moles of sodium acetate reduced by 0.002542 moles and the moles of acetic acid increased by 0.002542 moles.

no of moles of sodium acetate after the addition of HCl = 0.125853 - 0.002542

                                                                               = 0.123301 moles

resultant molarity of sodium acetate = 0.123301 mole/0.2012 L = [0.123301 / 0.2012] M

[since the total volume = 195 + 6.2 = 201.2 mL = 0.2012 L]

no of moles of acetic acid after the addition of HCl = 0.069147 + 0.002542

                                                                        = 0.071689 moles

resultant molarity of acetic acid = [0.071689 / 0.2012] M

Therefore [salt]/[acid] after addition of HCl = [0.123301/0.2012] / [0.071689/0.2012]

                                                             = 0.123301/0.071689

                                                             = 1.72

log [salt]/[acid] = log 1.72 = 0.2355

Therefore the resultant pH = 4.74 + 0.2355

                                      = 4.9755

The change in pH = 4.9755 - 5

                          = -0.0245

That is the pH is reduced by 0.0245.