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Equilibrium

Question
CBSEENCH11006539

Assuming complete dissociation, calculate the pH of the following solution:
0.002M KOH

Solution

0.002M KOH
As KOH is completely dissociated,  therefore, 
open square brackets OH to the power of minus close square brackets space equals space 0.002 space straight M space equals space 2 space cross times space 10 to the power of negative 3 end exponent straight M
Also comma space open square brackets straight H to the power of plus close square brackets space open square brackets OH to the power of minus close square brackets space equals space 1 space cross times space 10 to the power of negative 14 end exponent
therefore space space space space open square brackets straight H to the power of plus close square brackets space equals space fraction numerator 1 space cross times space 10 to the power of negative 14 end exponent over denominator 10 space straight H to the power of negative 1 end exponent end fraction space equals space fraction numerator 1 space cross times space 10 to the power of negative 14 end exponent over denominator 2 space cross times space 10 to the power of negative 3 end exponent end fraction
space space space space space space space space space equals space 5 space cross times space 10 to the power of negative 12 end exponent straight M
therefore space space pH space equals space minus log space open square brackets straight H to the power of plus close square brackets space equals space minus log space open square brackets 5 space cross times space 10 to the power of negative 12 end exponent close square brackets
space space space space space space space space space space equals space minus space open square brackets 0.6990 minus 12 close square brackets space equals space 11.301