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Question

Solutions A and B both containing strong bases have pH= 12 and pH=13 respectively. Both the solutions have an equal volume. The pH of the resultant solution obtained after mixing the two solutions is:
Take log(5.5)=0.74

A
13.74
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B
12.74
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C
11.26
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D
10.26
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Solution

The correct option is B 12.74
pH of solution A = 12
pOH=14pHpOH=2
We know that pOH=log[OH]
2=log[OH]
[OH]1st Solution=102 M

pH of solution B = 13
pOH=14pHpOH=1413=1pOH=1
Since, pOH=log[OH]
1=log[OH]
[OH]2nd Solution=101 M

Assuming both the solutions to be of 1 L
V1=V2=1 L

Total volume (Vf)=V1+V2

Vf=1+1=2 L

Cfinal=[OH]final=C1V1+C2V2V1+V2[OH]final=(102×1)+(101×1)2[OH]final=5.5×102 M

pOHmix=log[OH]finalpOHmix=log(5.5×102)pOHmix=(2log(5.5))=(20.74)pOHmix=1.26pHmix=141.26=12.74

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