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=14−pHpOH=2 We know that pOH=−log[OH−] 2=−log[OH−] [OH−]1stSolution=10−2M
pH of solution B = 13 pOH=14−pHpOH=14−13=1pOH=1 Since, pOH=−log[OH−] 1=−log[OH−] [OH−]2ndSolution=10−1M