Answer to Question #144497 in General Chemistry for martha

Question

sodium cyanide is the salt of a weak acid HCN. calculate the concentration of H3O+; OH-; HCN and Na+ in solution prepared by dissolving 10.8g of NaCN in enough water to make 5.00x 10^2 mL of solution at 25 degrees celsius

Answer 1

Mass of NaCN dissolve, W = 10.8g

Molar mass of NaCN, M = 49 g/mol

Volume of water, V = 5.00x 10^2 mL

Molar Concentration of NaCN,

C = (W×1000)/(49×5.00x 10^2) M

= 0.44 M

NaCN is a strong electrolyte complete dissociation give equimolar of Na+ and CN–

NaCN ------> Na+ + CN–

So, concentration of Na+ = 0.44 M

To calculate the concentration of H3O+, OH– and HCN , we have to consider the hydrolysis of NaCN in aqueous solution as,

CN– + H2O ---> HCN + OH–

The formula of degree of hydrolysis,

α =√(Kw/Ka×C)

Where ,

Kw = dissociation constant of water

= 10–14

Ka = dissociation constant of HCN

= 7.2×10^-10

C = Molarity of NaCN

= 0.44 M

Now,

Concentration,[OH–] = αC

Or, [OH–] = √(Kw×C/Ka)

Putting the values we get

Or, [OH–] =

√(10^-14×0.44/7.2×10^-10)

Or, [OH–] = 6.111×10^-5 M

From the above hydrolysis equilibrium Concentration,[HCN] = [OH–]

So, [HCN] = 6.111×10^-5 M

HCN is a weak acid

So, concentration

[H3O+] = √(Ka× concentration of acid,HCN)

Or, [H3O+] = √(Ka×[HCN])

Or, [H3O+] =

√(7.2×10^-10×6.111×10^-5)

Or, [H3O+] = 2.097×10^-7 M