Equilibrium - Result Question 45
####47. Equal volumes of three acid solutions of $pH 3,4$ and 5 are mixed in a vessel. What will be the $H^{+}$ ion concentration in the mixture?
[2008]
(a) $1.11 \times 10^{-4} M$
(b) $3.7 \times 10^{-4} M$
(c) $3.7 \times 10^{-3} M$
(d) $1.11 \times 10^{-3} M$
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Solution:
- (b) $[H_3 O]^{+}$for a solution having $pH=3$ is given by
$[H_3 O]^{+}=1 \times 10^{-3}$ moles/litre $[\therefore[H_3 O]^{+}=10^{-pH}]$
Similarly for solution having $pH=4$, $[H_3 O]^{+}=1 \times 10^{-4} moles /$ litre and for $pH=5$
$[H_3 O^{+}]=1 \times 10^{-5} moles /$ litre
Let the volume of each solution in mixture be 1L, then total volume of mixture solution
$ =(1+1+1) L=3 L $
Total $[H_3 O]^{+}$ion present in mixture solution $=(10^{-3}+10^{-4}+10^{-5})$ moles
Then $[H_3 O]^{+}$ion concentration of mixture solution
$ \begin{aligned} & =\frac{10^{-3}+10^{-4}+10^{-5}}{3} M=\frac{0.00111}{3} M \\ & =0.00037 M=3.7 \times 10^{-4} M . \end{aligned} $
