Question

Given the following information:

nitrous acid	HNO2	Ka = 4.5×10-4
hydrocyanic acid	HCN	Ka = 4.0×10-10

(1) Write the net ionic equation for the reaction that occurs when equal volumes of 0.242 M aqueous nitrous acid and sodium cyanide are mixed. It is not necessary to include states such as (aq) or (s).

_____	+	_____		_____	+	_____

(2) At equilibrium the _________(reactants/products) will be favored.

(3) The pH of the resulting solution will be _________(greater than/equal to/less than) seven.

Answer 1

(i) HNO2(aq) H+(aq) + NO2-(aq) ................. Ka1 = 4.5 x 10-4

(ii) HCN(aq) H+(aq) + CN-(aq) ................. Ka2 = 4.0 x 10-10

So,

(iii) CN-(aq) +  H+(aq) HCN(aq) ................... K = 1/Ka2

When HNO2 and CN-(from NaCN) are mixed, the net ionic equation will be (i) + (iii), i.e.,

Ans (1): HNO2(aq) + CN-(aq) HCN(aq) + NO2-(aq)

Keq of the above reaction = Ka1.K = Ka1/Ka2 = (4.5 x 10-4)/(4.0 x 10-10) = 1.125 x 106

Clearly, Keq 1.

Ans (2): Therefore, at equilibrium, PRODUCTS will be favored.

Concentration of HNO2 just after mixing = (0.242 M)/2 = 0.121 M

Concentration of CN- just after mixing = (0.242 M)/2 = 0.121 M

Reaction	HNO2(aq)	+	CN-(aq)		HCN(aq)	+	NO2-(aq)
Initial Conc.	0.121 M		0.121 M		0 M		0 M
Change in Conc.	-0.121 M		-0.121 M		+0.121 M		+0.121 M
New Conc.	0 M		0 M		0.121 M		0.121 M
Change in Conc.	+y M		+y M		-y M		-y M
Equilibrium Conc.	y M		y M		(0.121 - y) M		(0.121 - y) M

The reaction is made to go to completion first(since, Keq 1) and then allowed to come back to equilibrium stage to make our calculations easier and more precise.

Keq = ([HCN][NO2-])/([HNO2][CN-]) = 1.125 x 106

(0.121 - y)2/y2 = 1.125 x 106

(0.121 - y)/y = 1060.66

y = 1.14 x 10-4 M

Simultaneously, individual reaction equilibriums also hold.

Reaction	HCN(aq)		H+(aq)	+	CN-(aq)
New Conc.	0.121 M		0 M		0 M
Equilibrium Conc.	(0.121 - y) M		y M		y M

So, [H+] = y = 1.14 x 10-4 M

Now, pH = -log10([H+]) = -log(1.14 x 10-4) = 3.94

Ans (3): Therefore, pH = 3.94, pH of the resulting solution will be LESS THAN 7.