Question

Solve the following problems :
How many times concentrated will be the aqueous solution having pH $11.9$ as compared to aqueous solution having pH $8$ ?

Answer 1

For aqueous solution with pH $11.9$, pOH is $2.1$.

$pOH=-{\log }_{10}\left[O{H}^{-}\right]$

$2.1=-{\log }_{10}\left[O{H}^{-}\right]$

${\log }_{10}\left[O{H}^{-}\right]=-2.1$

${\log }_{10}\left[H_3{O}^{+}\right]=3–2.1–3$

${\log }_{10}\left[H_3{O}^{+}\right]=-3+0.9$

${\log }_{10}\left[H_3{O}^{+}\right]=+0.9$

$\left[H_3{O}^{+}\right]=anti\log \left(0.9\right)$

$\left[H_3{O}^{+}\right]=7.943\times {10}^{-3}M$

For aqueous solution with pH $8$, pOH is $6$.

$pOH=-{\log }_{10}\left[O{H}^{-}\right]$

$-{\log }_{10}\left[O{H}^{-}\right]=6$

${\log }_{10}\left[O{H}^{-}\right]=-6$

$\left[O{H}^{-}\right]=1\times {10}^{-6}M$ (${\log }_{a}b=m$ for $a^m=b$)

$\left[H_3{O}^{+}\right]$ in aquoeus solution having pH $11.9$ / $\left[H_3{O}^{+}\right]$

in aquoeous solution on having pH $8$

$=\frac{7.943\ast {10}^{-3}}{1\ast {10}^{-6}}=7943$

Thus, the aqueous solution with pH $8$ is $7943$ times more concentrated than the aqueous solution with pH $11.9$.