A 24.3 mL sample of 0.349 M dimethylamine , (CH 3 ) 2 NH , is titrated with 0.27
ID: 499242 • Letter: A
Question
A 24.3 mL sample of 0.349 M dimethylamine, (CH3)2NH, is titrated with 0.275 M hydrochloric acid.
(1) Before the addition of any hydrochloric acid, the pH is _____
(2) After adding 13.6 mL of hydrochloric acid, the pH is _____
(3) At the titration midpoint, the pH is _____
(4) At the equivalence point, the pH is _____
(5) After adding 45.9 mL of hydrochloric acid, the pH is _____
Equilibrium constants are found here: https://docs.google.com/document/d/1cNVOoEOVMqPSx8QdL8mc3zd8yFTwBx8VWd2MiDCDA1o/edit
Explanation / Answer
(CH3)2NH , pKb = 3.27
(1) Before the addition of any hydrochloric acid, the pH is
pH = 14- 1/2 {pKb -logC}
pH = 14 - 1/2 [3.27 -log0.349]
pH = 12.14
(2) After adding 13.6 mL of hydrochloric acid, the pH is
millimoles of B = 24.3 x 0.349 = 8.48
millimoles of H+ = 13.6 x 0.275 = 3.74
B + H+ ----------------> BH+
8.48 3.74 0
4.74 0 3.74
pH = 14 - (3.27 + log (3.74 / 4.74)
pH = 10.83
(3) At the titration midpoint, the pH is
pH = 10.73
(4) At the equivalence point, the pH is
pH = 5.77
(5) After adding 45.9 mL of hydrochloric acid, the pH is
pH = 1.23
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