wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
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
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
7.3×104 mol L1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3.7×103 mol L1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
7.3×105 mol L1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
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







flag
Suggest Corrections
thumbs-up
11
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon