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At 50 degree centrigade , Kw = 10^(-12) ...

At 50 degree centrigade , `K_w` = `10^(-12)` , pH of N/1000 KOH will be :

A

3

B

9

C

11

D

10

Text Solution

AI Generated Solution

To find the pH of a N/1000 KOH solution at 50 degrees Celsius where \( K_w = 10^{-12} \), we can follow these steps: ### Step 1: Understand the ionic product of water At 50 degrees Celsius, the ionic product of water (\( K_w \)) is given as \( 10^{-12} \). This means: \[ K_w = [H^+][OH^-] = 10^{-12} \] ...
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