ph of a solution of a strong acid is 5.0.What will be the ph of the solution obtained after dilluting the given solution to 100 times ?

Explanation

pH=5
or -log `[H^+]=5`
`therefore [H^+]=10^(-5)` M
On diluting 100 times , `[H^+]=10^(-5)/100=10^(-7)` M
From the formula, `pH=-log [H^+]` , its pH comes out to be 7 . It is not possible because an acidic solution cannot have pH=7. Since the solution is dilute ,
`[H^+]_"total"=[H^+]_"acid"+[H^+]_"water"`
Since HCl is strong acid
`[H^+]_"HCl"=1.0xx10^(-7)`
The concentration of `H^+` from ionization of water is equal to the `[OH^-]` from water
`[H^+]_(H_2O)=[OH^-]_(H_2O)` =x (say)
But `[H^+][OH^-]=1.0xx10^(-14)`
`(1xx10^(-7)+x)x=1.0xx10^(-14)`
`1xx10^(-7)x+x^2-1.0xx10^(-14)=0`
Solving we get `x=1.0xx10^(-7)`
`[H^+]=1.0xx10^(-7)+1.0xx10^(-7)`
`=2.0xx10^(-7)`
pH=-log `(2.0xx10^(-7))`
=7-log 2
=7-0.3010=6.6990