Question

Question 3

Sodium nitrite (NaNO₂) is a commonly used meat preservative.

a. Knowing that the K_a of nitrous acid (HNO₂) is 4.5 x 10^-4 M, calculate the pH of a 0.25 M sodium nitrite (NaNO₂) solution.

b. How many grams of sodium nitrite (NaNO₂) must be added to 100.0 mL of 0.150 M nitrous acid (HNO₂) to make a buffer solution with a pH of 3.50?

c. If 0.05 g of NaOH (s) are added to 100.0 ml of the buffer solution prepared in part (b), what is the new pH?

Answer 1

a)

Ionization equilibrium of HNO2

HNO2(aq) + H2O(l) <--------> H3O⁺(aq) + NO2^-(aq)

K_a = [H3O⁺] [NO2^-]/[HNO2] = 4.5 ×10^-4

at equilibrium

[HNO2] = 0.25 - x

[NO2^-] = x

[H3O⁺] = x

so,

x²/(0.25 - x) = 4.5 ×10^-4

solving for x

x = 0.01038

[H3O⁺] = 0.01038M

pH = -log[H3O⁺]

pH = -log(0.01038)

pH = 1.98

b)

pKa = -logKa

pKa of HNO2 = -log(4.5×10^-4)

pKa = 3.35

Henderson-Hasselbalch equation is

pH = pKa + log([A^-]/[HA])

3.50= 3.35 + log([NO2^-]/[HNO2])

log([NO2^-]/[HNO2]) = 0.15

[NO2^-]/[HNO2] = 1.412

moles of NO2^- = moles of HNO2 × 1.412

moles of HNO2 = ( 0.150mol/1000ml) ×100ml = 0.015mol

moles of NO2^- required = 0.015mol × 1.412 = 0.02118mol

mass of NaNO2 required = 0.02118mol × 69g/mol = 1.461g

Therefore,

Mass of NaNO2 required = 1.461g

c)

Number of moles of NaOH = 0.05g/40g/mol = 0.00125mol

Initial moles of NO2^- = 0.02118mol

Initial moles of HNO2 = 0.01500mol

NaOH react with weak acid HNO2

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

1:1 molar reaction

0.00125 moles of NaOH react with 0.00125moles of HNO2

After addition of NaOH

moles of HNO2 = 0.01500mol - 0.00125mol = 0.01375mol

moles of NO2^- = 0.02118mol + 0.00125mol = 0.02243mol

[HNO2] = ( 0.01375mol/ 100ml) ×1000ml = 0.1375M

[NO2^-] = ( 0.02243mol/100ml) × 1000ml = 0.2243M

Applying Henderson -Hasselbalch equation

pH = pKa + log([A^-]/[HA])

pH = 3.35 + log( 0.2243M/0.1375M)

pH = 3.35 + 0.21

pH = 3.56