An analytical chemist is titrating 209.0 ml of a 0.9400 M solution of HCH3CO2 wi

Question

An analytical chemist is titrating 209.0 ml of a 0.9400 M solution of HCH3CO2 with a 0.8100 M solution of KOH. The pKa of a acetic acid is 4.70. Calculate the pH of the acid solution after the chemist has added 260.3 ml of the KOH solution to it.

Round you answer to 2 decimal places.

??.all 100% 6:50 O ACIDS AND BASES Calculating the pH of... An analytical chemist is titrating 209.0 mL of a 0.9400 M solution of acetic acid (IICH3CO2 The p Ka of acetic acid is 4.70. Calculate the pH of the acid solution after the chemist has added 20 Note for advanced students: you may assume the final volume equals the initial volume of the solution added Round your answer to 2 decimal places. Explanation Check : 2018 McGraw Hill Eclucation. All Rights Rescrvedl. Terms of Usc | Privacy

Explanation / Answer

Moles = molarity *Volume in liter

moles of acid- 0.209*0.94= .19646

moles of base= 0.81*0.260 = .2106

So acid is limiting reagent

Moles of salt form = moles of acid consumed

Moles of salt= .19646

Moles of Koh remaining= .2106-.19646= .01414

Concentration of koh= .01414/(.209+.260)

=.0301M

Here OH-would be in solution so

OH- = .0301M

POh= 1.521

pH= 14-1.521= 12.47

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