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Question

500 mL of a strong acid solution of pH = 2 is mixed with 500 mL of another strong base solution of pOH = 3. The pH of resulting solution will be :
[Given, log(4.5)=0.65]

A
3.65
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B
2.35
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C
3.35
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D
2.65
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Solution

The correct option is B 2.35

pHsolution1=2 ; C1=[H+]=102pOHsolution2=3 ; C2=[OH]=103Cf=[H+]remains after neutralization V1=500 mL , V2=500 mLVf=V1+V2=1000 mLApplying:
C1V1C2V2=CfVfCf=C1V1C2V2Vf=(102× 500)(103× 500)1000 =4.5×103pH= log10[4.5×103]pH=30.65=2.35

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