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What will be the resultant pH when 200 m...

What will be the resultant pH when 200 mL of an aqueous solution of HCl (pH =2.0) is mixed with 300 mL of an aqueous solution of NaOH (pH =12.0) ?

Text Solution

Verified by Experts

The correct Answer is:
`11.31`
pH=2 means `[H^(+)]=10^(-2)M`
pH=12 means `[H^(+)]=10^(-12)` or `[OH^(-)]=10^(-2)M`
Thus, 200mL of `10^(-2)M HCl` are mixed with 300mL of `10^(-2)M NaOH`. After neutralisation NaOH left `=100mL` of `10^(-2)M`.
Total volume after mixing `=500mL`
`:.` In the final solution, after mixing,
`[OH^(-)]=(10^(-2))/(5)=2xx10^(-3)M`
or `[H^(+)]=(10^(-14))/(2xx10^(-3))=5xx10^(-12)`
`:.pH=-loh[H^(+)]=-log(5xx10^(-12))=12-0.69`
`=11.31`
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