Question

What is the pH of the solution formed when 30.0 g of potassium nitrite (KNO2) is dissolved in 1.00 L of water.

Answer 1

Mass of KNO₂ = 30.0 g

Molar mass of KNO₂ = 85.10379 g/mol

Moles of KNO₂ = mass of KNO₂/molar mass of KNO₂

= 30.0 g/85.10379 g/mol

= 0.353 mol

Concentration of KNO₂ = moles of solute/liter of solution

= 0.353 mol/1.00 L

= 0.353 M

pK_a of HNO₂ = 3.3

Therefore, pK_b of NO₂^- = (14 - 3.3) = 10.7 [ pK_a + pK_b = 14]

or, - logK_b = 10.7

or, K_b = 10^-10.7

= 2.0 x 10^-11

NO₂^- + H₂O HNO₂ + OH^-

Initial concentration (M)                                       0.353 0             0

Change in concentration (M)                                 -X                           X             X

Equilibrium concentration (M)                         (0.353 - X)                    X             X

Now,

K_a = [HNO₂][OH^-]/[NO₂^-]

or, 2.0 x 10^-11 = (X)(X)/(0.353 -X)

or, 2.0 x 10^-11 = (X)(X)/0.353 [(0.353 -X) 0.353 as X << 0.353]

or, X² = 7.06 x 10^-12

or, X = 2.66 x 10^-6

or, [OH^-] = 2.66 x 10^-6

or, -log[OH^-] = -log(2.66 x 10^-6) [taking - log function on both sides of the equation]

or, pOH = 5.58

or, 14 - pH = 5.58 [ pH + pOH = 14]

or, pH = 8.42

Hence, the pH of the solution = 8.42