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Question

Calculate the pH of a solution formed by mixing equal volume of two solutions A and B of a strong acid having pH=6 and pH=4 respectively ?
(Given: log 5.05 = 0.703)

A
4.29
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B
5
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C
5.29
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D
None of the above
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Solution

The correct option is A 4.29
pH of solution A = 6
we know that pH=log[H+]
6=log[H+]
[H+]=antilog(6)=106 molL1

pH of solution B = 4
we know that pH=log[H+]
4=log[H+]
[H+]=antilog(4)=104 molL1

Let's take case of 1 L solution of each.
total volume =1+1=2 L
Amount of H+ ion in 1 L of solution A = concentration× volume =106×1=106 mole
Similarly, Amount of H+ ion in 1 L of solution B =104×1=104 mole

Total amount of H+ in the solution formed by mixing solution A and B
=(106+104)=1.01×104 mole
Total concentration of H+=total amount of H+total volume=1.01×1042=5.05×105
pH=log[H+]=log(5.05×105)
pH=log(5.05)+5
pH=0.703+5
pH=4.297

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