Question

In: Chemistry

Consider the reaction HCN(aq) H+(aq) +CN-(aq). the equilibrium constant is 6.2*10^-10. If you place 0.4 mols...

Consider the reaction HCN(aq) H+(aq) +CN-(aq). the equilibrium constant is 6.2*10^-10. If you place 0.4 mols of HCN in a 2.0 liter flask, what is the equilbrium of CN-? so far ive got to x=x(0.2)(6.2*10^-10)

Expert Solution

[HCN] = mol of HCN / volume

= 0.4 mols / 2.0 L

= 0.2 mols

Lets write the dissociation equation of HCN

HCN -----> H+ + CN-

0.2 0 0

0.2-x x x

Kc = [H+][CN-]/[HCN]

Kc = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kc = x*x/(c)

so, x = sqrt (Kc*c)

x = sqrt ((6.2*10^-10)*0.2) = 1.114*10^-5

[CN-] = x = 1.114*10^-5

Answer: 1.11*10^-5

queen_honey_blossom answered 2 years ago

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