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100mL of " 0.1 M NaOH" solution is titra...

`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))`

A

7

B

7.2

C

7.4

D

6.8

Text Solution

Verified by Experts

The correct Answer is:
2
`{:(,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`
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Knowledge Check

  • 100 ml of 0.3 M HCl solution is mixed with 200 ml of 0.3 M H_(2)SO_(4) solution. What is the molariyt of H^(+) in resultant solution ?

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