An 2.2507 g sample of oxalic acid (mw=90.03 g/mol) is dissolved in water to make a 250.0 ml solution. given that ka1 = 6.5x10-2, ka2 = 6.1x10-5 ,what is the ph of the solution? (

Chemical reaction (dissociation) 1: C₂O₄H₂(aq) ⇄ C₂O₄H⁻(aq) + H⁺(aq).
Chemical reaction (dissociation) 2: C₂O₄H⁻(aq) ⇄ C₂O₄²⁻(aq) + H⁺(aq).
c(C₂O₄H⁻) = c(H⁺) = x.
m(C₂O₄H₂) = 2.2507 g.

n(C₂O₄H₂) = m(C₂O₄H₂) ÷ M(C₂O₄H₂).

n(C₂O₄H₂) = 2.2507 g ÷ 90.03 g/mol.

n(C₂O₄H₂) = 0.025 mol.

V(C₂O₄H₂) = 250.0 mL = 0.25 L.

c(C₂O₄H₂) = 0.025 mol / 0.25 L = 0.1M.

Ka₁ = 6.5·10⁻².
Ka₁ = c(C₂O₄H⁻) · c(H⁺) / c(C₂O₄H₂).
0.065 = x² / (0.1 M - x).
Solve quadratic eqaution: x = c(H⁺) = 0.0544 M.
pH(oxalic acid) = -log(0.0544 M) = 1.26.

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).