Question

Determine pH of ${10}^{-8}$ M HCL solution

Answer 1

HCl is a strong acid. Therefore it dissolve completely.

$HCl⟶H^{+}C{l}^{-}$

At $t=0{10}^{-8}M$

At $t=\infty 0{10}^{-8}{10}^{-8}$

$H^{+}$ from $H_{2}O$ is also considered

$H_{2}O⟶H^{+}+O{H}^{-}$

$xx$

$+{10}^{-8}$

$K_{eq}={10}^{-14}$

$K_{eq}=\left[H^{+}\right]\left[O{H}^{-}\right]$

${10}^{-14}=(x+{10}^{-8})\left(x\right)$

${10}^{-14}=x^2+{10}^{-8}x$

Multiplying with ${10}^8$ on the both sides.

${10}^{-6}={10}^8{x}^2+x$

${10}^8{x}^2+x-{10}^{-6}=0$

$l=\frac{-1_{-}^{+}\sqrt{1+4\times {10}^8\times {10}^{-6}}}{2\times {10}^8}$

$x=\frac{-1_{-}^{+}\sqrt{1+400}}{2\times {10}^8}$

$\frac{-1+20.025}{2\times {10}^8}=\frac{19.025\times {10}^{-8}}{2}$

$=9.513\times {10}^{-8}$

$\therefore$ total $\left[H^{+}\right]={10}^8+9.513\times {10}^{-8}$

$=10.513\times {10}^{-8}$

$=10.513\times {10}^{-8}$

$pH=-lo{g}_{10}\left[H^{+}\right]$

$-lo{g}_{10}(10.513\times {10}^{-8})$

$8-log\left(10.513\right)$

$pH=8-1.022=6.978$