Ionic EquilibriumHard

Question

Consider a weak base ' B ' of ${pK}_{b} = 5.699$. ' x ' mL of 0.02 M HCl and ' y ' mL of 0.02 M weak base ' B ' are mixed to make 100 mL of a buffer of pH 9 at $25^{\circ}C$. The values of ' $x$ ' and ' $y$ ' respectively are:

[Given: $log2 = 0.3010,log3 = 0.4771,log5 = 0.699$ ]


Options

A.

11.1

88.9

B.

42.7

57.3

C.

14.3

85.7

D.

85.7

14.3

Solution

HCl $$+ B \rightarrow {BH}^{+} + {Cl}^{-}$$
0.02 M 0.02 M
x ml y ml
$$t_{f}$$ 0 $$0.02y - 0.02x$$ 0.02 x

$${pOH = {pK}_{b} + log\left\lbrack \frac{\text{~Salt~}}{\text{~Base~}} \right\rbrack }{5 = 5.699 + log\left\lbrack \frac{\text{~Salt~}}{\text{~Base~}} \right\rbrack }{\frac{x}{y - x} = \frac{1}{5} }{6x = y }{7x = 100 }{x = \frac{100}{7}ml }{\& y = \frac{600}{7}ml}$$

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