`100mL` of `" 0.1 M NaOH"` solution is titrated with `100mL` of `"0.5 M "H_(2)SO_(4)` solution. The `pH` of the resulting solution is `: (` For `H_(2)SO_(4), K_(a1)=10^(-2))`

Options

A. 7
B. 7.2
C. 7.4
D. 6.8

Answer

2

Explanation

`{:(,NaOH,+,H_(2)SO_(4),,rarr,NaHSO_(4),+,H_(2)O),("initial m mole",10,,5,,,0,,),("reaction",5,,0,,,5,,),(,NaOH,+,NaHSO_(4),,rarr,Na_(2)SO_(4),+,H_(2)O),(,5,,5,,,0,,),(,0,,0,,,5,,):}`
`pH=7+(1)/(2)[2+log.(5)/(200)]=7+(1)/(2)[2+log5-log200]=7+(1)/(2)[2+0.7-0.3-2]=7.2`