Question

A solution is formed by adding 3.60 grams of solid barium cyanide, Ba(CN)₂, to 250.0 ml of 0.170M sodium cyanide, NaCN. Calculate the pH of the resulting solution. You may assume that the barium cyanide dissolve completely and that the change in volume is negligible with the addition of the solid.

Answer 1

Given:

Mass of barium cyanide (Ba(CN)₂) = 3.60 grams

The volume of sodium cyanide (NaCN) = 250.0 mL = 0.250 L

Molarity of sodium cyanide (M) = 0.170 M

Solution:

Sodium cyanide (NaCN) is a basic salt which obtained by reaction of a weak acid (HCN) and strong base (NaOH)

NaCN (aq) → Na⁺ (aq) + CN^- (aq)

Barium cyanide dissolves completely and change in volume is negligible.

Ba(CN)₂ → Ba²⁺ + 2CN^-

The molar mass of barium cyanide = 189.362 g/mol

Find out the number of mole of cyanide obtained from barium cyanide.

Mole = mass/molar mass

The number of mole of barium cyanide = 3.60 g/189.362 g/mol = 0.0190 mol

One mole of barium cyanide gives two moles of cyanide

0.0190 mol of barium cyanide gives (0.0190 * 2)/1 = 0.038 mol of cyanide

The total number of mole of cyanide in solution = number of mole of cyanide from barium cyanide + number of mole of cyanide from sodium cyanide.

Molarity = Number of mole / volume

Number of mole = Molarity * volume

Number of mole = 0.170 mol/L * 0.250 L = 0.0425 mol

The total number of mole of cyanide in solution = 0.038 mol + 0.0425 mol = 0.0805 mol

The concentration of cyanide = moles total / volume

The concentration of cyanide = 0.0805 mol / 0.250 L = 0.322 M

Hydrolysis reaction: CN^- (aq) + H₂O (l) HCN (aq) + OH ^- (aq)

K_a of HCN = 6.2 * 10^-10

K_a = dissociation constant of an acid

	CN- (aq)	HCN (aq)	OH- (aq)
Initial concentration	0.322 M	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.322– x	x	x

Dissociation constant of an acid is the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants each raised to the power of their coefficients.

Where x is very small, so x is negligible in 0.322 – x

x = 1.4129 * 10^-5 M

[OH^-] = x = 1.4129 * 10^-5 M

pH (potent of hydrogen) is defined as negative logarithms of concentration of hydrogen ion.

pOH = - log [OH^-]

pOH = - log (1.4129 * 10^-5)

pOH = - log (1.4129) + 5log10

pOH = - 0.1501 + 5

pOH = 4.8499

pOH = 4.85

pH + pOH = 14

pH = 14 – pOH

pH = 14 - 4.85

pH = 9.15

The pH of the resulting solution is 9.15