Question

Ionic product of water is 4* 10^-13 at a particular temperature (t^0 C).The neutral point based on a pH scale at t^0 is

Answer 1

Answer on Question#41125-Chemistry-Physical chemistry

Question.

Ionic product of water is $4^{*}10^{\wedge}-13$ at a particular temperature $(t^{0}C)$. The neutral point based on a pH scale at $t^{0}$ is

Solution.

Ionic product of water is:

$$K_{W} = \left[ H_{3} O^{+} \right] \left[ O H^{-} \right]$$

Neutral point condition:

$$\left[ H_{3} O^{+} \right] = \left[ O H^{-} \right]$$ $$K_{W} = \left[ O H^{-} \right]^{2}$$ $$4 * 10^{-13} = \left[ O H^{-} \right]^{2}$$ $$\left[ H_{3} O^{+} \right] = \left[ O H^{-} \right] = 0.63 * 10^{-6}$$ $$pH = - \log \left( \left[ H_{3} O^{+} \right] \right) = 6.2$$

Answer: Neutral point pH = 6.2

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