At 277 K, degree of dissociation water is 1× 10- 7 % . The value of ionic product of water is

Question

At 277 K, degree of dissociation water is 1× 10- 7 % . The value of ionic product of water is (A) 3.0× 10- 14 (B) 3.085× 10- 15 (C) 1× 10- 16 (D) 1× 10- 14

Tardigrade Admin · Accepted Answer

The molar concentration of water =(1000 × 1/18)=55.5 text M α , the degree of dissociation =(1 × 1 0- 7/100) =1× 10- 9 <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-foyhtnrn5yof.png" /> Kw=Cα × Cα or C2α 2 Kw=(55.5)2(1 × 10-9)2=3.085 × 10-15

Q. At 277 K, degree of dissociation water is $1 \times 1 0^{- 7} %$ . The value of ionic product of water is

$3.0 \times 1 0^{- 14}$

$3.085 \times 1 0^{- 15}$

$1 \times 1 0^{- 16}$

$1 \times 1 0^{- 14}$

The molar concentration of water

$= \frac{1000 \times 1}{18} = 55.5 M$

$α$ , the degree of dissociation $= \frac{1 \times 1 0 ^{- 7}}{100}$

$= 1 \times 1 0^{- 9}$

$K_{w} = C α \times C α$ or $C^{2} α^{2}$

$K_{w} = (55.5)^{2} (1 \times 1 0^{- 9})^{2} = 3.085 \times 1 0^{- 15}$

Q. At 277 K, degree of dissociation water is 1×10−7% . The value of ionic product of water is

3.0×10−14

3.085×10−15

1×10−16

1×10−14