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Q.

A $0.052 M$ solution of benzoic acid, $C_{6} H_{5} COOH$ , is titrated with a strong base. What is the [ $H^{+}$ ] of the solution one-half way to the equivalence point?
Equilibrium Constant, $K_{a}$
$C_{6} H_{5} COOH$ $6.3 \times 10^{- 5}$

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a

$6.3 \times 10^{- 5} M$

b

$1.8 \times 10^{- 3} M$

c

$7.9 \times 10^{- 3} M$

d

$2.6 \times 10^{- 2} M$

answer is A.

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Detailed Solution

At half way to equivalence point,

$[C_{6} H_{5} {COO}^{-}] = [C_{6} H_{5} COOH]$

Therefore,

$K_{a} = [H^{+}] \Rightarrow [H^{+}] = 6.3 \times 10^{- 5}$

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A 0.052 M solution of benzoic acid, C6H5COOH, is titrated with a strong base. What is the [H+] of the solution one-half way to the equivalence point?EquilibriumConstant, KaC6H5COOH6.3×10−5