Answer:
I think it's B. The ability of water molecules to adhere to the surfaces of objects
Given:
Diprotic weak acid H2A:
Ka1 = 3.2 x 10^-6
Ka2 = 6.1 x 10^-9.
Concentration = 0.0650 m
Balanced chemical equation:
H2A ===> 2H+ + A2-
0.0650 0 0
-x 2x x
------------------------------
0.065 - x 2x x
ka1 = 3.2 x 10^-6 = [2x]^2 * [x] / (0.065 - x)
solve for x and determine the concentration at equilibrium.
Answer: C = 0.014M
Explanation:
From n= m/M= CV
m =43.5 M= 148, V=850ml
43.5/148= C× 0.85
C= 0.35M
Applying dilution formula
C1V1=C2V2
C1= 0.35, V1= 25ml, C2=?, V2= 600ml
0.35× 25 = C2× 600
C2= 0.014M
Given the solubility of strontium arsenate is 0.0480 g/l . we have to convert it into mol/L by dividing it over molar mass (540.7 g/mol)
Molar solubility = 0.0480 / 540.7 = 8.9 x 10⁻⁵ mol/L
Dissociation equation:
Sr₃(AsO₄)₂(s) → 3 Sr²⁺(aq) + 2 AsO₄³⁻(aq)
3 s 2 s
Ksp = [Sr²⁺]³ [AsO₄³⁻]²
= (3s)³ (2s)²
= 108 s⁵
Ksp = 108 (8.9 x 10⁻⁵) = 5.95 x 10⁻¹⁹
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