SATHEE: Equilibrium - Result Question 59

Equilibrium - Result Question 59

####61. At $100^{\circ} C$ the $K_{w}$ of water is 55 times its value at $25^{\circ} C$ . What will be the $p H$ of neutral solution? $(\log 55 = 1.74)$

[NEET Kar. 2013]

(a) 6.13

(b) 7.00

(d) 5.13

Show Answer

Solution:

(a) $K_{w}$ at $25^{\circ} C = 1 \times 10^{- 14}$

At $25^{\circ} C$

$K_{w} = [H^{+}] [O H^{-}] = 10^{- 14}$

At $100^{\circ} C$ (given)

$K_{w} = [H^{+}] [O H^{-}] = 55 \times 10^{- 14}$

$∵$ for a neutral solution

$[H^{+}] = [O H^{-}]$

$∴ [H^{+}]^{2} = 55 \times 10^{- 14}$

or $[H^{+}] = (55 \times 10^{- 14})^{1 / 2}$

$∵ p H = - \log [H^{+}]$ On taking $\log$ on both side

$\begin{aligned} - \log [H^{+}] = - \log (55 \times 10^{- 14})^{1 / 2} \\ p H = \frac{1}{2} (- \log 55 + 14 \log 10) \\ p H = 6.13 \end{aligned}$

Calculation of $p O H$ in this question: value of $p H$ and $p O H$ must be same for a neutral solution. Thus, $p O H = 6.13$ , also $p H + p O H = - \log (55 \times 10^{- 14})$

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Equilibrium - Result Question 59

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