Question

suppose that 100ml of 0.100 M Bacl2 is mixed with 100 ml of 0.150M Na2SO4.
what is pH of resulting solution ? when Ksp= 1.5 ×10^-9.
I really want to understand systems of this equation. are they mixed in water? so the sulfate ion reacts with water?? procuding bisulfate ion with hydroxide ion??

To be clear, My question is ... 1) when they are mixed together, they exist in Baso4 (s), sodium ion, chloride ion, and water? 2) after precipitate formed, the remaining sulfate ions reacts with water resulting in Hso4^- and OH^-.

Am I understanding correct? if not, please explain me correct explanation. I'd be very much appreciated if you answer me more simple way.
Thank you so much!

Answer 1

Yes, you are absolutely correct.

When these two salts are mixed, BaSO4(s) is formed, leaving behind Na⁺(aq), Cl^-(aq) and excess SO₄^2- in the solution.

Initial millimoles of Ba²⁺(aq) = M*V = 0.100 mol/L * 100 mL = 10 millimol

Initial millimoles of SO₄^2-(aq) = M*V = 0.150 mol/L * 100 mL = 15 millimol

The chemical equation for the formation of BaSO4(s) is:

---Ba²⁺(aq) + SO₄^2-(aq) --------> BaSO₄(s)

I: 10 ---------- 15 ---------------------- 0

C: - 10 -------- - 10 -------------------- +10

E: (10 - 10), (15 - 10) ----------------- 10

Moles of SO₄^2-(aq) remained in the solution = 15 - 10 = 5 millimol

Total volum of solution, Vt = 100+100 = 200 mL

=> [SO₄^2-(aq)] = 5 millimol / 200 mL = 0.025 M

Now SO₄^2-(aq) undergoes hydrolysis to form HSO₄^-(aq) and OH^-(aq):

--- SO₄^2-(aq) + H2O < ------>  HSO₄^-(aq) + OH^-(aq) : Kb = Kw / Ka2 = 9.77*10^-13

I: 0.025 M ------------------------ 0 --------------- 10^-7 M  

C: - X ------------------------------ +X -------------- +X

E: (0.025 - X) -------------------- X ---------------- (10^-7+ X)

Kb = 9.77*10^-13 = [HSO₄^-(aq)]*[OH^-(aq)] / [SO₄^2-(aq)]

=> 9.77*10^-13 = [X*(10^-7+ X)] / (0.025 - X)

Since X << 0.025

=>  9.77*10^-13 = [X*10^-7+ X²)] / 0.025

=> X*10^-7+ X² - 2.44*10^-14 = 0

Solving the above quadratic equation gives: X = 1.14*10^-7

=> [OH^-(aq)] = (10^-7+ X)

=> [OH^-(aq)] = 10^-7+ 1.14*10^-7 = 2.14*10^-7  

=> pOH = - log[OH^-(aq)] = - log(2.14*10^-7) = 6.67

=> pH = 14 - pOH = 14 - 6.67 = 7.33 (Answer)