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Edit View History People Window Help xe C| secure | https://www.awh.a ALEKS Learn-Abert Yeboa O ACIOS, BASES AND AQUEOUS EQUILIBRIA Calculating the pH of a weak acid titrated with a strong base An analytical chemist is trating 234.5 mL of a 0.9900 M solution of propionic acid (HC2H,CO2) with a 0.2100 M solution of KOH. The pK, of propionic acid is 4.89. Calculate the pH of the acid solution after the chemist has added 405.7 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places. PH Check

Explanation / Answer

we have below equation to be used:

pKa = -log Ka

4.89 = -log Ka

log Ka = -4.89

Ka = 10^(-4.89)

Ka = 1.288*10^-5

we have:

Molarity of HC2H5CO2 = 0.99 M

Volume of HC2H5CO2 = 234.5 mL

Molarity of KOH = 0.21 M

Volume of KOH = 405.7 mL

mol of HC2H5CO2 = Molarity of HC2H5CO2 * Volume of HC2H5CO2

mol of HC2H5CO2 = 0.99 M * 234.5 mL = 232.155 mmol

mol of KOH = Molarity of KOH * Volume of KOH

mol of KOH = 0.21 M * 405.7 mL = 85.197 mmol

We have:

mol of HC2H5CO2 = 232.155 mmol

mol of KOH = 85.197 mmol

85.197 mmol of both will react

excess HC2H5CO2 remaining = 146.958 mmol

Volume of Solution = 234.5 + 405.7 = 640.2 mL

[HC2H5CO2] = 146.958 mmol/640.2 mL = 0.2296M

[C2H5CO2-] = 85.197/640.2 = 0.1331M

They form acidic buffer

acid is HC2H5CO2

conjugate base is C2H5CO2-

Ka = 1.288*10^-5

pKa = - log (Ka)

= - log(1.288*10^-5)

= 4.89

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 4.89+ log {0.1331/0.2296}

= 4.65

Answer: 4.65

Feel free to comment below if you have any doubts or if this answer do not work

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