Question

Calculate the ph of a buffer solution prepared by dissolving 30g of na2co3 in 500 ml of an aqueous solution containing 150 ml of 1m hcl. ka for hco-3 = 5.63 x 10-11

Answer 1

Hydrochloric acid reacts with Sodium carbonate to neutralise the base as follows:

$Na_2CO_3 + HCl  >  NaCl + H_2CO_3$

$Moles of Na2CO3 = m /M = 30 /105 = 0.28 mol$

$Moles of HCl = V(L) * M = (0.15) * 1 = 0.15 mol$

$Mole of Na2CO3 left = 0.28 – 0.15 = 0.13 mol$

$Concentration of left Na2CO3 = 0.13 / (0.500 + 0.15) M$

$Concentration of Salt = 0.15 / (0.500 + 0.15) M$

$pH of buffer = pKa + log [salt] /[Base]$

$pKa = - log [ka] = - log [5.63 * 10^-^1^1 ] = 10.32$

$pH = 10.32 + log [0.15] /[0.13]  = 10.32 + 0.11 = 10.43$

Thus, pH of buffer is 10.43

Answer 2

30 g Na₂CO₃ in 500 ml solution ≡ [(30/106)/(500/1000)] M Na₂CO₃ ≡ 0.57 M Na₂CO₃

Initial H⁺ concentration coming from HCl = 150 X 1 = 150 milimoles

Initial CO₃⁻ concentration coming from Na₂CO₃ = 500 X 0.57 = 285 milimoles

Thus after reacting, excess CO₃⁻ left = (285 - 150) = 135 milimoles

Concentration of salt = 150 / (500+150) = 0.23 milimoles/ml

Concentration of Na₂CO₃ = 135 / (500+150) = 0.21 milimoles/ml

Now, kₐ = 5.63 X 10⁻¹¹

or, pkₐ = -log [kₐ] = -log [5.63 X 10⁻¹¹] = 10.23

pH = pkₐ + log [salt / base]

or, pH = 10.23 + log [0.23/0.21]

or, pH = 10.23 + 0.04

or, pH = 10.27

pH of the buffer is 10.27.