Question

An analytical chemist is titrating 195.2mL of a 0.8100M solution of benzoic acid HC6H5CO2 with a 0.7200M solution of NaOH. The pKa of benzoic acid is 4.20. Calculate the pH of the acid solution after the chemist has added 87.44mL of the NaOH solution to it.

Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added.

Round your answer to 2 decimal places.

An analytical chemist is titrating 195.2 mL of a 0.8100 M solution of benzoic acid (HCHCO2) with a 0.7200 M solution of NaOH. The pK, of benzoic acid is 4.20. Calculate the pH of the acid solution after the chemist has added 87.44 mL of the NaOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added. El Round your answer to 2 decimal places. B p 3 83 cm

Answer 1

Given:

M(HC6H5CO2) = 0.81 M

V(HC6H5CO2) = 195.2 mL

M(NaOH) = 0.72 M

V(NaOH) = 87.44 mL

mol(HC6H5CO2) = M(HC6H5CO2) * V(HC6H5CO2)

mol(HC6H5CO2) = 0.81 M * 195.2 mL = 158.112 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.72 M * 87.44 mL = 62.9568 mmol

We have:

mol(HC6H5CO2) = 158.112 mmol

mol(NaOH) = 62.9568 mmol

62.9568 mmol of both will react

excess HC6H5CO2 remaining = 95.1552 mmol

Volume of Solution = 195.2 + 87.44 = 282.64 mL

[HC6H5CO2] = 95.1552 mmol/282.64 mL = 0.3367M

[C6H5CO2-] = 62.9568/282.64 = 0.2227M

They form acidic buffer

acid is HC6H5CO2

conjugate base is C6H5CO2-

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.2+ log {0.2227/0.3367}

= 4.021

Answer: 4.02