10 is titrated with a 0.0228 Msolu A 50.00-mL solution of hydrochlonis solution

Question

10 is titrated with a 0.0228 Msolu A 50.00-mL solution of hydrochlonis solution of 0.0350 M/trimethylamine (, -6.5x a) 9.8I b) 8.16 c) 3.02 d) 4.19 97 ochloric id as the titrant What is the phl at the equivalence point? (K- 1.0 x 10) e) 5.84 Ans: e Algorithm: Yes Chapter/Section: 15.4 Difficulty: Difficult Keyword 1: Chemistry Keyword 6: titration of a weak base Keyword 2: general chemistry Keyword 3: acids and bases Keyword 4: solutions of a weak acid or base with another solute Keword5: acid-base titration curve by a strong acid

Explanation / Answer

At the equivalence point, volume of HCl needed = 50.00 * 0.0350 / 0.0228 = 76.75 ml.

so total volume of solution = (50.00 + 76.75) = 126.75 ml.

molarity of salt = 0.050 * 0.0350 * 1000 / 126.75 = 0.0138 M

here salt hydrolysis occurs.

PKb = -log Kb = - log (6.5 * 10^-5) = 4.187

PH = 0.5 * PKw - 0.5 * PKb - 0.5 * log C

PH = 0.5 * 14 - 0.5 * 4.187 - 0.5 * log (0.0138)

PH = 7 - 2.09 + 0.93

PH = 5.84

so option e) 5.84 is the answer.

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