[4 marks Questions]

Question

How many times would be concentrated aqueous solution having pH = 11.9 be more basic as compared to aqueous solution having pH = 8?

Vidyadip · Accepted Answer

pOH of basic aqueous solution having pH = 8 will be 14 - 8 = 6.
$\therefore\left[ OH ^{-}\right]=1 \times 10^{-6} M$
Similarly, pOH of basic aqueous solution having pH 11.9 will be $14-11.9=2.1$.
Now, $pOH =-\log _{10}\left[ OH ^{-}\right]$
$\begin{array}{l}
\therefore 2.1=-\log _{10}\left[OH^{-}\right] \\
\therefore-(2.1)=\log _{10}\left[OH^{-}\right] \\
\therefore \log _{10}\left[OH^{-}\right]=\overline{3} .9 \\
\therefore\left[OH^{-}\right]=\text {antilog } \overline{3} .9000=7.943 \times 10^{-3} M
\end{array}$
Now,
$\frac{\left[ OH ^{-}\right] \text {in aqueous solution having pOH } 2.1}{\left[ OH ^{-}\right] \text {in aqueous solution having } pOH 6}$
$=\frac{7.943 \times 10^{-3}}{1 \times 10^{-6}}=7.943 \times 10^3=7943$
Thus, the solution having pH 11.9 will have 7943 times more basicity than solution having pH 8.

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