Answer to Question #339733 in Chemistry for MARJ

Question

2.   A carbonate buffer of pH 10 is needed by a chemist. How many grams of Na₂CO₃ is needed to dissolve in 1.5L of 0.2M NaHCO₃ to make the buffer? K_a of HCO₃^- = 4.7 x 10^-11

Answer 1

Given:

pH = 10

[NaHCO₃] = 0.2 M

Liters of solution = 1.5 L

K_a of HCO₃⁻ = 4.7×10⁻¹¹

Solution:

The Henderson-Hasselbalch equation is commonly used to calculate the pH of a buffer solution from the concentration of the buffer components:

where pK_a is the acid dissociation constant and [base] and [acid] are the base and acid concentrations, respectively.

In this specific reaction, the base is Na₂CO₃ and the acid is NaHCO₃.

Therefore,

10 = −log(4.7×10⁻¹¹) + log([Na₂CO₃] / 0.2)

−0.3279 = log([Na₂CO₃] / 0.2)

0.47 = [Na₂CO₃] / 0.2

[Na₂CO₃] = 0.094

Thus, the molarity of Na₂CO₃ is 0.094 M

Molarity = Moles of solute / Liters of solution

Therefore,

Moles of Na₂CO₃ = Molarity of Na₂CO₃ × Liters of solution = (0.094 M) × (1.5 L) = 0.141 mol

Moles = Mass / Molar mass

The molar mass of Na₂CO₃ is 106 g/mol

Therefore,

Mass of Na₂CO₃ = (0.141 mol Na₂CO₃) × (106 g Na₂CO₃ / 1 mol Na₂CO₃) = 14.946 g Na₂CO₃

Mass of Na₂CO₃ = 14.95 g

Answer: 14.95 grams of Na₂CO₃ is needed to dissolve