Equilibrium Ques 58
58. At $100^{\circ} C$ the $K_w$ of water is 55 times its value at $25^{\circ} C$. What will be the $pH$ of neutral solution? $(\log 55=1.74)$
[NEET Kar. 2013]
(a) 6.13
(b) 7.00
(c) 7.87
(d) 5.13
Show Answer
Solution:
- (a) $K_w$ at $25^{\circ} C=1 \times 10^{-14}$
At $25^{\circ} C$
$K_w=[H^{+}][OH^{-}]=10^{-14}$
At $100^{\circ} C$ (given)
$K_w=[H^{+}][OH^{-}]=55 \times 10^{-14}$
$\because$ for a neutral solution
$ [H^{+}]=[OH^{-}] $
$\therefore \quad[H^{+}]^{2}=55 \times 10^{-14}$
or $[H^{+}]=(55 \times 10^{-14})^{1 / 2}$
$\because pH=-\log [H^{+}]$ On taking $\log$ on both side
$ \begin{aligned} & -\log [H^{+}]=-\log (55 \times 10^{-14})^{1 / 2} \\ & pH=\frac{1}{2}(-\log 55+14 \log 10) \\ & pH=6.13 \end{aligned} $
Calculation of $pOH$ in this question: value of $pH$ and $pOH$ must be same for a neutral solution. Thus, $pOH=6.13$, also $pH+pOH=-\log (55 \times 10^{-14})$
BETA
