Question

Sodium carbonate (2.4134 g) is dissolved in enough deionized water to give a solution with a total volume of 250.0 mL. What is the pH of the resulting solution? Hint: For carbonic acid, pKa1 = 6.351 and pKa2 = 10.329. What is the equilibrium concentration of H2CO3 in the solution? calculate the value of alpha HCO3-

Answer 1

Ok we have sodium carbonate that will dissociate into

Na₂CO₃ 2Na⁺ + CO₃^-2

we can find the moles of sodium carbonate.

n= 2.4134 g / 105,988 g/mol = 0.02277 mol of Na₂CO₃

1 mol of Na₂CO₃ will produce 1 mol of CO₃ =  0.02277 mol of CO₃

[CO₃ ] =  0.02277 mol / 0.250 L = 0.0911 M

The carbonate ion is the conjugate base of the weak acid HCO− 3 (K = 10−10.7), so this solution will be alkaline. Except in very dilute solutions, the pH should be sufficiently high to preclude the formation of any significant amount of H2CO3, so we can treat this problem as a solution of a monoprotic weak base:

CO₃²⁻ + H₂O OH⁻ + HCO ₃ ⁻

Kb = [OH⁻]*[HCO₃⁻ ]/ [CO3 ²⁻ ] = Kw/ Ka = 10^{−14 /}10^−10.3 = 10^−3.7 = 1.99x10^-4

Neglecting the OH− produced by the autoprotolysis of water, we make the usual assumption that

[OH⁻] = [HCO₃⁻] , and thus

Kb = [OH⁻]² / 0.0911− [OH⁻]

Assuming that 0.0911 - OH^- = 0.0911

1.99x10^{-4 *} 0.0911 =  [OH⁻]²

[OH^-] = 4.26x10^-3

pOH = -log [OH^-]

pOH = 2.37

pH = 14 - pOH

pH = 11.62