Question

What mass of sodium nitrite must be added to enough water to make 350.0 mL of a solution with pH = 8.40? [Ka(HNO2) = 5.6 × 10–4]

Answer 1

8.5187g

Explanation

pOH = 14 - pH

pOH = 14 - 8.40

pOH = 5.60

pOH = - log[OH^-]  

-log[OH^-] = 5.60

[OH^-] = 2.51 × 10^-6M

NO2^- ion partly hydrolysed by water

NO2^- (aq) + H2O(l) <--------> HNO2(aq) + OH^-(aq)

K_b = [HNO2][OH^-]/[NO2^-]

K_b = K_w/K_a

where,

K_w = ionic product of water, 1.00 × 10^-14

K_a = dissocian constant of HNO2 , 5.6 × 10^-4

K_b = 1.00 ×10^-14/5.6 ×10^-4 = 1.786 × 10^-11

[OH^-] = [HNO2]

[NO2^-]_eq​​​​​_uillibrium = [HNO2][OH^-]/K_b

[NO2^-]_equillibrium = (2.51×10^-6M)²/1.786×10^-11

[NO2^-]_equillibrium = 0.352749

[NO2^-]_equillibrium = [NO2^-]_initial - (2.51×10^-6)

[NO2^-]_initial = [NO2^-]_equillibrium + 2.51 × 10^-6

[NO2^-]_initial = 0.352749M + 0.00000251M

[NO2^-]_initial = 0.35275M

moles of NO2 required for 350.0ml = (0.35275mol/1000ml)×350.0ml = 0.12346mol

moles of NaNO2 required = 0.12346mol

mass of NaNO2 required = 0.12346mol × 69g/mol = 8.5187g