The correct option is A 3.7×10−4 mol L−1
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=3⇒3=−log[H+]⇒[H+]=antilog(−3)=10−3 molL−1
For II nd solution pH=−log[H+]
For Ist solution pH=4⇒4=−log[H+]⇒[H+]=antilog(−4)=10−4 molL−1
For III rd solution pH=−log[H+]
For Ist solution pH=5⇒5=−log[H+]⇒[H+]=antilog(−5)=10−5 molL−1
But, as the volume of each solution is 1 litre so,
using the formula,
Moles=Molarity×Volume
For Ist solution Moles of H+=10−3 molL−1×1L=10−3 mol
For II nd solution Moles of H+=10−4 molL−1×1L=10−4 mol
For III rd solution Moles of H+=10−5 molL−1×1L=10−5 mol
After the mixing is done,
Total volume = 3 L
Total moles (=10−3+10−4+10−5)
=111×10−5 mol
So, for: For final solution [H+]=111×10−53
=37×10−5
=3.7×10−4 molL−1
hence, the H+ ion concentration of the mixture is 3.7×10−4 mol L−1