Calculating the pH of a weak acid titrated with a strong b An analytical chemist

Question

Calculating the pH of a weak acid titrated with a strong b An analytical chemist is titrating 87.0 ml. of a 0.4200 M solution of butanoic acid (HC,H,Co2) with a 0.8800 M solution of KOH. The pK, of butanoic acid is 4.82. Calculate the pH of the acid solution after the chemist has added 44.0 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.

Explanation / Answer

Balanced equation:

C3H7COOH (aq) + KOH(aq) -> H2O(l) + C3H7COOK(aq)

87.0 ml of 0.4200M C3H7COOH

So, moles of C3H7COOH = 0.4200M x (87.0 /1000L) = 0.03654 moles

44.0 ml of 0.8800M KOH

So, moles of KOH = (0.8800M x (44/1000) = 0.03872 moles

Reaction stoichiometry shows, there is 1: 1 molar ratio between C3H7COOH and KOH

0.03654 Moles of butanoic acid is neutralizes 0.03654 moles of KOH

After adding 44.0 ml of 0.8800M KOH to 87.0 ml of 0.4200M C3H7COOH, KOH neutralizes all acid. There are few more moles of KOH

Excess moles of KOH = 0.03872 moles - 0.03654 moles = 0.00218 moles

Volume of solution = 87.0 ml + 44.0 ml = 131 ml

Molarity of KOH = 0.03872 moles/(131/1000)L = 0.2956 M

Now calculate the pH of a 0.2956 M solution of KOH.

KOH(s) ---> K+(aq) + OH¯

100% of the KOH molecules dissociate in solution.

The [OH¯] of a strong acid is equal to the concentration of the base.

[OH¯] = 0.2956 M

pOH = - log (0.2956) = 0.53

pH = 14.00 - pOH

pH = 14.00 – 0.53 = 13.47

pH of acid solution after addition of 44.0 ml of 0.8800M KOH is 13.47

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