Question

a)You have an aqueous solution of a weak monoprotic base with a volume of 300 mL and a molarity of 0.150 M. you add to it 0.075 moles of its weak conjugate acid. You then add 6.00 g of barium hydroxide. Assuming the volume change is negligible what is the pOH of the resulting solution? b)You prepare a buffer with 0.500 moles of a weak monoprotic acid and 0.150 moles of its weak conjugate base. If you add 200 mL of a 0.485 M HCl to the. Mixture, what would the resulting pH be? The Ka of the weak acid was 3.92 x10^-4

Answer 1

a) Molar mass of barium hydroxide, Ba(OH)₂ = 171.34 g/mol.

6.00 g Ba(OH)₂ = (6.00 g)*(1 mole/171.34 g) = 0.0350 mole Ba(OH)₂.

Now Ba(OH)₂ ionizes as

Ba(OH)₂ (aq) -------> Ba²⁺ (aq) + 2 OH^- (aq)

Let HA be the weak monoprotic acid. We have 300.00 mL of 0.150 M HA; therefore, mole(s) of HA = (300.00 mL)*(1 L/1000 mL)*(0.150 M) = 0.0450 mole.

HA reacts with Ba(OH)₂ to give BaA₂ where A^- is the conjugate base of HA. The acid-base reaction is

Ba(OH)₂ (aq) + 2 HA (aq) --------> BaA₂ (aq) + 2 H₂O (l)

As per the stoichiometric equation,

1 mole Ba(OH)₂ = 2 moles HA.

Therefore, 0.0450 mole HA = (0.0450 mole)*(1 mole Ba(OH)₂/2 mole HA) = 0.0225 mole Ba(OH)₂.

Mole(s) of Ba(OH)₂ left unreacted = (0.0350 – 0.0225) mole = 0.0125 mole.

Now Ba(OH)₂ ionizes as

Ba(OH)₂ (aq) -------> Ba²⁺ (aq) + 2 OH^- (aq)

Due to the 1:2 nature of ionization, 0.0125 mole Ba(OH)₂ = 2*0.0125 mole = 0.0250 mole OH^-.

[OH^-] = (0.0250 mole)/[(300.00 mL)*(1 L/1000 mL)] = 0.08333 mol/L.

pOH = -log [OH^-] = -log (0.08333) = 1.0792 ≈ 1.08 (ans).