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Question: At 100°C, the $K_w$ of water is 55 times its value at 25°C(log 55 = 1.74). T"he pH of a neutral queo...

At 100°C, the KwK_w of water is 55 times its value at 25°C(log 55 = 1.74). T"he pH of a neutral queous solution of 100°C is

A

5.13

B

6.13

C

7.0

D

7.87

Answer

6.13

Explanation

Solution

The ionic product of water, KwK_w, at 25°C is 1.0×10−141.0 \times 10^{-14}.

Given that at 100°C, the KwK_w of water is 55 times its value at 25°C. So, Kw(100∘C)=55×Kw(25∘C)K_w (100^\circ C) = 55 \times K_w (25^\circ C) Kw(100∘C)=55×(1.0×10−14)K_w (100^\circ C) = 55 \times (1.0 \times 10^{-14}) Kw(100∘C)=55×10−14K_w (100^\circ C) = 55 \times 10^{-14}

For a neutral aqueous solution, the concentration of hydrogen ions ([H+][H^+]) is equal to the concentration of hydroxide ions ([OH−][OH^-]). Also, Kw=[H+][OH−]K_w = [H^+][OH^-]. Since [H+]=[OH−][H^+] = [OH^-] for a neutral solution, we can write: Kw=[H+]2K_w = [H^+]^2 Therefore, [H+]=Kw[H^+] = \sqrt{K_w}

Substitute the value of KwK_w at 100°C: [H+]=55×10−14[H^+] = \sqrt{55 \times 10^{-14}} [H+]=55×10−14[H^+] = \sqrt{55} \times \sqrt{10^{-14}} [H+]=55×10−7[H^+] = \sqrt{55} \times 10^{-7} M

Now, calculate the pH using the formula pH=−log⁡[H+]pH = -\log[H^+]: pH=−log⁡(55×10−7)pH = -\log(\sqrt{55} \times 10^{-7}) Using the logarithm properties log⁡(AB)=log⁡A+log⁡B\log(AB) = \log A + \log B and log⁡An=nlog⁡A\log A^n = n \log A: pH=−[log⁡(55)+log⁡(10−7)]pH = -[\log(\sqrt{55}) + \log(10^{-7})] pH=−[log⁡(551/2)−7]pH = -[\log(55^{1/2}) - 7] pH=−[12log⁡(55)−7]pH = -[\frac{1}{2}\log(55) - 7]

We are given log⁡55=1.74\log 55 = 1.74. pH=−[12(1.74)−7]pH = -[\frac{1}{2}(1.74) - 7] pH=−[0.87−7]pH = -[0.87 - 7] pH=−[−6.13]pH = -[-6.13] pH=6.13pH = 6.13

The pH of a neutral aqueous solution at 100°C is 6.13.