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# An analytical chemist is titrating 160.3mL of a 1.100M solution of acetic acid HCH3CO2 with a...

An analytical chemist is titrating 160.3mL of a 1.100M solution of acetic acid HCH3CO2 with a 0.4100M solution of NaOH. The pKa of acetic acid is 4.70. Calculate the pH of the acid solution after the chemist has added 114.1mL 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 160.3 mL of a 1.100 M solution of acetic acid (HCH2CO,) with a 0.4100 M solution of N2OH. The p K, of acetic acid is 4.70. Calculate the pH of the acid solution after the chemist has added 114.1 mL of the N2OH 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. pH X 18 Ar

Given:

M(HCH3CO2) = 1.1 M

V(HCH3CO2) = 160.3 mL

M(NaOH) = 0.41 M

V(NaOH) = 114.1 mL

mol(HCH3CO2) = M(HCH3CO2) * V(HCH3CO2)

mol(HCH3CO2) = 1.1 M * 160.3 mL = 176.33 mmol

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

mol(NaOH) = 0.41 M * 114.1 mL = 46.781 mmol

We have:

mol(HCH3CO2) = 176.33 mmol

mol(NaOH) = 46.781 mmol

46.781 mmol of both will react

excess HCH3CO2 remaining = 129.549 mmol

Volume of Solution = 160.3 + 114.1 = 274.4 mL

[HCH3CO2] = 129.549 mmol/274.4 mL = 0.4721M

[CH3CO2-] = 46.781/274.4 = 0.1705M

They form acidic buffer

acid is HCH3CO2

conjugate base is CH3CO2-

use:

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

= 4.7+ log {0.1705/0.4721}

= 4.258

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