Answer: The value of \(K_w \text{ at } 25^oC \text{ is }1\times 10^{-14}\)
Explanation:
\(H_2O\rightleftharpoons H^++OH^-\)
\(K_{eq}=\frac{[H^+]\times [OH^-]}{[H_2O]}\)
\(K_{eq}\times[H_2O]=K_w=[H^+]\times [OH^-]\)
Since, water is neutral with value pH of 7 at \(25^oC\). So, the concentration of \([H^+] and [OH^-]\) will be equal.
\(pH=7=-log[H^+]=\) so, the value of \(H^+\) will be
\([H^+]=1\times 10^{-7}mol/L=[OH^-]\)
The value of \(K_w\) at \(25^o C\) will be :
\(K_w=[H^+]\times [OH^-]=1\times 10^{-7}\times 1\times 10^{-7}=1\times 10^{-14}\)
