At 25° C, acid dissociation constant of HCN is 4.9× 10- 10M. Calculate the % increase in degree of dissociation on diluting HCN solution from 0.1M to 0.001M. Represent the answer as sum of digits till you get single digit answer.

Question

At 25° C, acid dissociation constant of HCN is 4.9× 10- 10M. Calculate the % increase in degree of dissociation on diluting HCN solution from 0.1M to 0.001M. Represent the answer as sum of digits till you get single digit answer. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Ka=CÎ±2 α i=√(4 . 9 × 10- 10/0 . 1)=7× 10- 5 α F=√(4 . 9 × 10- 10/10- 3)=√49 × 10- 8=7× 10- 4 (α F - α i/α i)× 100=(7 × 10- 4 - 7 × 10- 5/7 × 10- 5)× 100=900 %

Q. At 25∘C, acid dissociation constant of HCN is 4.9×10−10M. Calculate the % increase in degree of dissociation on diluting HCN solution from 0.1M to 0.001M. Represent the answer as sum of digits till you get single digit answer.

Ka​=Cα2 αi​=0.14.9×10−10​​=7×10−5 αF​=10−34.9×10−10​​=49×10−8​=7×10−4 αi​αF​−αi​​×100=7×10−57×10−4−7×10−5​×100=900%

Q. At $2 5^{\circ} C,$ acid dissociation constant of $H CN$ is $4.9 \times 1 0^{- 10} M .$ Calculate the $%$ increase in degree of dissociation on diluting $H CN$ solution from $0.1 M$ to $0.001 M .$ Represent the answer as sum of digits till you get single digit answer.