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Question

Calculate pH of mixture of (400 mL,1200M Ba(OH)2)+(400 mL,150M HCl)+(200 mL of water)
log 2=0.3010

A
2.4
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B
7
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C
more than 7
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D
4.5
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Solution

The correct option is A 2.4
Given, VBa(OH)2=VHCl=400 mL, [Ba(OH)2]=1200M, [HCl]=150M, VH2O=200 mL
Number of H+ ions from acid N1V1=400×150=8
Number of OH ions from base N2V2=400×1200×2=4
here, N1V1>N2V2 so solution will be acidic in nature.
so, [H+]=N1V1N2V2Total volume
[H+]=(84)400+400+200
[H+]=4×103
so, pH=log [H+]=log 4×103
pH=32 log 2=2.4

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