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.

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6