Q.

The pH of a 0.01 M weak acid HX (Ka=4×10-10) is found to be 5. Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6.  The new concentration of the diluted weak acid is given as x×10-4 M.  The value of x is ______ (nearest integer).             [2025]


Ans.

(25)

New concentration of H+([H+]new)=10-pH(new)M=10-6M 

KaCnew=[H+]new

4×10-10Cnew=10-6

4×10-10Cnew=10-12 

Cnew=25×10-4M

Note: In the question it should be 0.25 M weak acid instead of 0.01 M weak acid. However, this data is not needed to solve the question.