Question

freezing point of a 0.1M formic acid aqueous solution is −0.20460C−0.20460C. Find equilibrium constant of the reaction - 
HCOO−(aq.)+H2O(ℓ)⇌HCOO(aq.)+OH−(aq.)HCOO−(aq.)+H2O(ℓ)⇌HCOO(aq.)+OH−(aq.)
Given : Kf(H2O)=1.86K−kgmole−1Kf(H2O)=1.86K−kgmole−1
Assume solution to be very dilute​

Answer 1

The equilibrium constant is 1.1×10⁻³

Given,

change in freezing point of formic acid = - 0.2046

cryoscopic constant = 1.86

molarity= .1

To Find,

equlibrium constant

Solution,

here, molarity = molalty = .1

ΔT= kf×m×i

i = ΔT/m×kf

i=0.2o46/0.1×1.86

= 1.1

i = 1-α+nα

where i  vant hoff factor, α is degreee of dissocaition and

i= 1-α+2α

1.1= 1+α

α= o.1 =  10%

so, the initial amount of HCOO is 0.1 and  10 % is decomposed into HCOOH and OH

so finally HCOO⁻ is .09 and OH⁻and HCOO⁻ is .01

the equilibrium constant is k = HCOOH×OH⁻/HCOO⁻

=.01×.01/.09

= 1.1×10⁻³ = equilibrium constant

#SPJ1

Answer 2

GIVEN:

• Molarity of (aq)formic acid = 0.1M
• Freezing point of Formic acid = -0.2046°c
• Cryoscopic constant = 1.86k-Kg/mole

TO FIND:

• The equilibrium constant of the reaction :

HCOO−(aq.)+H2O(ℓ)⇌HCOO(aq.)+OH−

SOLUTIONS:

We can simply find the value of the equilibrium constant as under:

Formula to calculate freezing point depression of a solution :

∆T = Kf × b × i

Putting the values in the above formula we get.

i = b × Kf / ∆T

= 0.1 × 1.86 / 0.2046

i = 1.1

Also, we know

i = an + (1-a)

where,

i = van 't Hoff factor

i = van 't Hoff factora = degree of dissociation

i = van 't Hoff factora = degree of dissociationn = the number of ions formed from one formula unit of the substance

here, In the above equation, we know that number of ions formed is 2

i = a×2 + (1-a)

1.1 = 2a (1-a)

a = 0.1 or,

a = 10%.

So, The initial amount of aqueous solution of formic acid is 0.1 m and 10% is decomposed into HCOO- and OH+.

So finally we have,

HCOO- = 0.9

OH+. = 0.1

HCOOH- = 0.1

APPLYING THE FORMULA:

EQUILIBRIUM CONSTANT

K = 0.1× 0.9 × 0.1

=

$1.1 \times {10}^{ - 3}$