Question

What is the pH of $2\times {10}^{-8}MHCl$ solution ?
(Take $log\left(11\right)=1.04$)

• A. 4.5
• B. 5.96
• C. 6.96
• D. 5.5

Answer 1

The correct option is C 6.96

$HCl\rightleftharpoons H^{+}+C{l}^{-}$
$[H^{+}{]}_{fromHCl}=2\times {10}^{-8}$
Let x be the $[H^{+}{]}_{ion}$ from $H_{2}O$
$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}(x+2\times {10}^{-8})x\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}(x+2\times {10}^{-8})x={10}^{-14}(x+2\times {10}^{-8})x-{10}^{-14}=0x^2+2\times {10}^{-8}x-{10}^{-14}=0x=\frac{-2\times {10}^{-8}\pm \sqrt{(2\times {10}^{-8}{)}^2+4\times {10}^{-14}}}{2}x=\frac{-2\times {10}^{-8}+(2\times {10}^{-7})}{2}x=\frac{1.8\times {10}^{-7}}{2}x=9\times {10}^{-8}pH=-{\log }_{10}(9\times {10}^{-8}+2\times {10}^{-8})pH=-\log [11\times {10}^{-8}]pH=8-1.04=6.96$