wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Calculate the pH of the solution formed by mixing equal volume of two solutions A and B at 25 oC having pH=10 and pH=12 repectively ?
(Given log (5.05) = 0.703)

A
11.7
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
10.29
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of the above
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
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)=104 mol L1

pH of solution B = 12
pOH of the solution = 14 -12 = 2
we know that pOH=log[OH]
2=log[OH]
[OH]=antilog(2)=102 mol L1

Assuming volume of 1 L solution of each.
total volume =1+1=2 L
Amount of OH ion in 1L of solution A = concentration× volume =104×1=104 mol
Similarly, Amount of OH ion in 1 L of solution B =102×1=102 mol

Total amount of OH in the solution formed by mixing solution A and B
=(104+102)=1.01×102 mol
Total concentration of OH=total amount of OHtotal volume=1.01×1022=5.05×103 mol L1
pOH=log[OH]=log(5.05×103)
pOH=log(5.05)+3
pOH=0.703+3
pOH=2.3
so pH=142.3=11.7

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon