Calculate the pH of the solution formed by mixing equal volume of two solutions A and B at 25oC having pH=10 and pH=12 repectively ?
(Given log (5.05) = 0.703)
A
11.7
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B
10.29
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C
12
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D
None of the above
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Solution
The correct option is A 11.7 pH of solution A = 10
then pOH = 14 - 10 =4
we know that pOH=−log[OH−] 4=−log[OH−] [OH−]=antilog(10)=10−4molL−1
pH of solution B = 12
pOH of the solution = 14 -12 = 2
we know that pOH=−log[OH−] 2=−log[OH−] [OH−]=antilog(2)=10−2molL−1
Assuming volume of 1L solution of each. ∴ total volume =1+1=2L
Amount of OH− ion in 1L of solution A = concentration× volume =10−4×1=10−4mol
Similarly, Amount of OH− ion in 1 L of solution B =10−2×1=10−2mol
Total amount of OH− in the solution formed by mixing solution A and B =(10−4+10−2)=1.01×10−2mol
Total concentration of OH−=totalamountofOH−totalvolume=1.01×10−22=5.05×10−3molL−1 pOH=−log[OH−]=−log(5.05×10−3) pOH=−log(5.05)+3 pOH=−0.703+3 pOH=2.3
so pH=14−2.3=11.7