# Question 83440

Mar 31, 2017

My Choice of option is (1) $p H = 6.95$

#### Explanation:

Let $c$ M be the concentration of ${H}^{+}$ in the acid solution of

$p H = 6$

So

$- {\log}_{10} c = 6$

$\implies {\log}_{10} c = - 6$

$\implies c = {10}^{-} 6$ M

If the solution is 100 times diluted then new concentration of ${H}^{+}$ in respect of acid will be $\left[{H}^{+}\right] = {10}^{-} 8$ M

If we take negative 10 base logarithm of this $\left[{H}^{+}\right] = {10}^{-} 8$ M to calculate the pH then $p H = - {\log}_{10} \left({10}^{-} 8\right) = 8$ This means the solution becomes basic (as $p H > 7$) but this is not feasible.

An approximate calculation

if we add concentration of ${H}^{+}$ due to ionization of water ${10}^{-} 7$ M with the concentration of ${H}^{+}$ from acid then total concentration of ${H}^{+} = {10}^{-} 8 + {10}^{-} 7 = {10}^{-} 7 \times 1.1$ M

So $p H = - {\log}_{10} \left({10}^{-} 7 \times 1.1\right) = 7 - {\log}_{10} 1.1 \approx 6.95$



More correct approach

Let water on ionisation contributes ${H}^{+} \mathmr{and} O {H}^{-}$ each in $x$ M
So
$\left[{H}_{\text{water"^+]=[OH_"water}}^{-}\right] = x$ M
As the ionic product of water${k}_{w} = {10}^{-} 14$ is a constant
then

([H_"acid"^+]+[H_"water"^+])xx[OH_"water"^-]=10^-14

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

$\implies {x}^{2} + {10}^{-} 8 x - {10}^{-} 14 = 0$

on solving we get

$x \approx 9.5 \times {10}^{-} 8$

So in acid solution

$\left[{H}_{\text{total}}^{+}\right] = {10}^{-} 8 + 9.5 \times {10}^{-} 8 = 10.5 \times {10}^{-} 8 = 1.05 \times {10}^{-} 7$

This gives

$p H = - {\log}_{10} \left[{H}_{\text{total}}^{+}\right] \approx 6.98$