Question

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 ? $(log 55 = 1.74)$

• A. 7
• B. 7.87
• C. 5.13
• D. 6.13

Answer 1

The Correct Option is D

To solve this problem, we need to understand the relationship between the ion product of water (\(K_w\)) and temperature, and how it affects the pH of water. 

1. At \(25^\circ C\), the ion product of water (\(K_w\)) is approximately \(\text{1} \times \text{10}^{-\text{14}}\).
2. At \(100^\circ C\), the problem states that \(K_w\) is 55 times its value at \(25^\circ C\), i.e., \(K_w = 55 \times 10^{-14}\).
3. For a neutral solution, \([\text{H}^+] = [\text{OH}^-]\) and \(K_w = [\text{H}^+][\text{OH}^-] = [\text{H}^+]^2\). Therefore, \([\text{H}^+] = \sqrt{K_w}\).
4. Calculate the value of \([\text{H}^+]\): \([\text{H}^+] = \sqrt{55 \times 10^{-14}} = \sqrt{55} \times 10^{-7}\).
5. Find the pH using the formula: \(\text{pH} = -\log_{10}[\text{H}^+]\).
6. Therefore, \(\text{pH} = -\log_{10}(\sqrt{55} \times 10^{-7})\) which simplifies to: \(\text{pH} = -\left(\dfrac{1}{2} \log_{10}(55) + \log_{10}(10^{-7})\right)\).
7. Substituting the given value, \(\log_{10}(55) = 1.74\): \(\text{pH} = -\left(\dfrac{1}{2} \times 1.74 - 7\right) = -\left(0.87 - 7\right)\).\)
8. Therefore, the pH at \(100^\circ C\) for a neutral solution is: \(\text{pH} = 6.13\).

Hence, the correct answer is 6.13, which matches the correct option provided.