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Vera_Pavlovna [14]
3 years ago
15

What is the equilibrium constant of pure water at 25°C?

Chemistry
2 answers:
ANEK [815]3 years ago
6 0
The equilibrium reaction, causes the water dissociation constant, Kw, is 1.01 × 10-14<span> at 25 °C. That is because every H</span>+<span> (H</span>3O+) ion these forms accompanied by the formation of an OH-<span> ion, are the concentrations of these ions and in pure water the same thing can be calculated from </span>Kw<span>. 

HOPED THIS HELP OUT ;)



</span>
Charra [1.4K]3 years ago
5 0

Answer:The equilibrium constant of pure water is K_{eq}=1\times 10^{-14}

Explanation:

H_2O+H_2O\rightleftharpoons H_3O^++OH^-

K_{eq}=\frac{[H_3O^+][OH^-]}{[H_2O][H_2O]}

K_i=K_{eq}\times [H_2O]}=\frac{[H_3O^+][OH^-]}{[H_2O]}

K_i=\text{Ionization constant of water}

Since,the water is found to be poorly ionized ,concentration of pure water practically remains the same. So, concentration of water can be combined with K_i to give new constant known as ionic product of water that is K_w.

K_w=K_i\times [H_2O]=[H_3O^+][OH^-]

In pure water:

[H_3O^+]=1\times 10^{-7} M]=[OH^-]

K_w=1\times 10^{-14}M^2

K_w=K_{eq}=1\times 10^{-14}

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