Question

Hydrocyanic acid will set up the following equilibrium when dissolved in water:

HCN(aq) + H₂O(l) <--> CN^-(aq) +H₃O⁺(aq)

a) calculate the pH of a .123 M solution of hydrocyanic acid

ANSWER: pH = 5.01

b) To 500mL of the solution in part (a) is added 5.00g of solid NaCN. Calculate the pH of new solution. Assume no volume change has occured.

c) If one were to add 10.00mL of a .100M NaOH solution to the solution prepared in part (b), what would the new pH be? Assume that the volume change is negligble and can be ignored.

Answer 1

a) Molar concentration of the acid HCN, c = 0.123 M

Since HCN is a weak acid, its H⁺ ion concentration is given by the relation; [H⁺] = (c Ka)^1/2

Ka of HCN = 6.2 x 10^-10

Thus the concentration of H⁺ ion in 0.123 M HCN solution = (c Ka)^1/2

= ( 0.123 X 6.2 x 10^-10)^1/2

= 8.73 X 10^-6 mol L^-1

Thus pH = - log [H⁺] = - log (8.73 X 10^-6 ) = 5.05

b) Volume of the solution = 500 mL = 0.5 L

Upon addition of NaCN, the solution will become buffer solution

Mass of NaCN added = 5.00g

Number of moles of NaCN  = mass of NaCN/molar mass of NaCN = 5.00 g/ 49 g/mol = 0.102 mol

So, [NaCN] = 0.102 mol / 0.5 L = 0.204 M

Then according to Hendereson-Hasselbalch equation pH pf the buffer solution;

pH = pKa + log ([salt] / [ acid])

= - log Ka + log ([NaCN] / [ HCN])

= - log (6.2 x 10^-10) + log (0.204 / 0.123)

= 9.21 + log (1.66)

= 9.21 + 0.22 = 9.43