Question

A 1.045 L sample of HCl gas at 25∘C and 732.0 mm Hg was absorbed completely in an aqueous solution that contained 7.086 g of Na2CO3 and 270.0 g of water. What is the pH of the solution?

Answer 1

We know that PV=nRT

Now 732 mm Hg = 0.9631 atm

R (gas constant)= 0.08205L atm K−1 mol−1

T = 298K

n= 0.9631 ×1.045/0.08205×298

n= 0.04116 moles of HCl

0.04116 moles of HCl dissolved in 270 g of water

Now density of water is 1000gm/L so, 270g of water = 0.27L of water

We can calculate molarity of HCl as follows,

0.04116 moles of HCl/0.27 L of water = 0.1524M

Molarity of HCl = 0.152M

Similarly molarity of Na2CO3 will be =(7.086gm/105.988)/0.27L = 0.2476M

The reaction between sodium carbonate and hydrochloric acid takes place in two stages

Na₂CO₃(aq) + HCl(aq) → NaHCO₃(aq) + NaCl(aq)-----------------------------------1

NaHCO₃(aq) + HCl(aq) → NaCl(aq) + CO₂(g) + H₂O(l)-----------------------------2

Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + CO₂(g) + H₂O(l)----------------------(Addition of equation 1 and 2)

So we can see from final equation that we need two moles of HCl to react completely with one mole of Na2CO3

Therefore [moles of Na2CO3]-[2*(moles of HCL)]= 0.057 moles of remaining unreacted HCl

Now this will dissociate completely to give H+ and Cl- ions

Now pH= -log [H+]

pH = -log 0.057

pH= 1.2