Question

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?

• A. $3.7\times {10}^{-3}M$
• B. $1.11\times {10}^{-3}M$
  
• C. $1.11\times {10}^{-4}M$
• D. $3.7\times {10}^{-4}M$

Answer 1

The correct option is D $3.7\times {10}^{-4}M$

$\because pH=-log\left[H^{+}\right]or\left[H^{+}\right]={10}^{-5}$
For sol. 1, $\left[H^{+}\right]={10}^{-3}$
For sol. 2, $\left[H^{+}\right]={10}^{-4}$ and
For sol. 3, $\left[H^{+}\right]={10}^{-5}$
Total $\left[H^{+}\right]={10}^{-3}+{10}^{-4}+{10}^{-5}$
$={10}^{-3}(1+1\times {10}^{-1}+1\times {10}^{-2})$
$={10}^{-3}(\frac{1}{1}+\frac{1}{10}+\frac{1}{100})$
$={10}^{-3}\left(\frac{100+10+1}{100}\right)$
$=1.11\times {10}^{-3}$
Now $H^{+}$ ion concentration in mixture of equal volume of these acid solution
$=\frac{1.11\times {10}^{-3}}{3}=3.7\times {10}^{-4}M$