To solve this problem, we first need to calculate the molar solubility of Ba(IO3)2 at 25°C using the Ksp value given.
Ba(IO3)2 (s) ⇌ Ba2+ (aq) + 2IO3- (aq)
The Ksp expression for this reaction is:
Ksp = [Ba2+][IO3-]^2
Ksp = 1.57 x 10^-9
Since 1 mol of Ba(IO3)2 produces 1 mol of Ba2+ and 2 mol of IO3-, the molar solubility of Ba(IO3)2 can be represented as "x".
So, the equilibrium concentrations of Ba2+ and IO3- ions will be "x" and "2x", respectively.
Ksp = x * (2x)^2
1.57 x 10^-9 = 4x^3
Solving for x:
x = (1.57 x 10^-9 / 4)^(1/3)
x ≈ 2.81 x 10^-4 M
Now, we can calculate the grams of Ba(IO3)2 that can be dissolved in 250 ml of water (0.25 L) at this concentration.
Molar mass of Ba(IO3)2 = 137.33 + 2(126.9 + 3(16)) = 487 g/mol
Number of moles of Ba(IO3)2 = concentration * volume = 2.81 x 10^-4 mol/L * 0.25 L = 7.03 x 10^-5 mol
Mass of Ba(IO3)2 = number of moles * molar mass = 7.03 x 10^-5 mol * 487 g/mol = 0.034 g
Therefore, the correct answer is:
0.034 g
How many grams of Ba(IO3)2 ( 487g/mol) can be dissolved in 250 ml of water at 25 C? Ksp = 1.57 * 10-9
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1 point
0.089 g
0.178 g
1.78 g
0.89 g
1 answer