Home
Class 12
CHEMISTRY
When 2.5 mL of 2//5M weak monoacidic bas...

When `2.5 mL` of `2//5M` weak monoacidic base `(K_(b) = 1 xx 10^(-12) at 25^(@)C)` is titrated with `2//15 M HCI` in water at `25^(@)C` the concentration of `H^(o+)` at equivalence point is `(K_(w) = 1 xx 10^(-14) at 25^(@)C)`

A

`3.7xx10^(-13) M`

B

`3.2xx10^(-7) M`

C

`3.2xx10^(-2) M`

D

`2.7xx10^(-2) M`

Text Solution

Verified by Experts

The correct Answer is:
D
Let 'v' ml of HCl is used reach equivalence point
`2.5xx 2/5= Vxx2/15 implies V=7.5ml`
Number of moles of salt formed =1m mole
Final total volume =2.5 + 7.5 = 10ml
Final conc. of salt`=1/10 = 0.1M`
`K_(h)=K_(w)/K_(b)=10^(-14)/10^(-12) = 10^(-2) implies K_(h)=(Ch^(2))/(1-h)` (we cannot assume (1-h) approx. 1)
Solve quadratic to find 'h', here, h=0.27
`[H^(+)] = Ch=0.1xx0.27=2.7xx10^(-2)M`
Promotional Banner

Similar Questions

Explore conceptually related problems

10 mL of 0.1 M tribasic acid H_(3)A is titrated with 0.1 M NaOH solution. What is the ratio of ([H_(3)A])/([A^(3-)]) at 2^(nd) equivalence point ? Given K_(1)=7.5xx10^(-4), K_(2)=10^(-8), K_(3)=10^(-12) .

Ksp of M(OH)_(2) is 5 xx 10^(-16) at 25^(0)C . The pH of its saturated solution at 25^(0)C is

The molarity of NH_(3) of pH = 12 at 25^(@)C is (K_(b) = 1.8 xx 10^(-5))

Calculate pOH of 0.1 M aq. Solution of weak base BOH (K_(b)=10^(-7)) at 25^(@)C.

Degree of dissociation of 0.1 molar acetic acid at 25^(@)C (K_(a) = 1.0 xx 10^(-5)) is