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Question

Equal volumes of three acid solutions of pH 3, 4, and 5 are mixed in a vessel. What will be the H+ ion concentration of the mixture ?
antilogb(x)=bx

A
3.7×104 mol L1
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B
7.3×104 mol L1
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C
3.7×103 mol L1
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D
7.3×105 mol L1
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Solution

The correct option is A 3.7×104 mol L1
Three solutions of equal volume are mixed.
Let's consider, initial volume of each solution= 1L
we know, pH=log[H+]
For Ist solution pH=33=log[H+][H+]=antilog(3)=103 molL1

For II nd solution pH=log[H+]
For Ist solution pH=44=log[H+][H+]=antilog(4)=104 molL1

For III rd solution pH=log[H+]
For Ist solution pH=55=log[H+][H+]=antilog(5)=105 molL1

But, as the volume of each solution is 1 litre so,
using the formula,
Moles=Molarity×Volume
For Ist solution Moles of H+=103 molL1×1L=103 mol
For II nd solution Moles of H+=104 molL1×1L=104 mol
For III rd solution Moles of H+=105 molL1×1L=105 mol

After the mixing is done,
Total volume = 3 L
Total moles (=103+104+105)
=111×105 mol

So, for: For final solution [H+]=111×1053
=37×105
=3.7×104 molL1
hence, the H+ ion concentration of the mixture is 3.7×104 mol L1







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