The ksp of Ba(IO3)2 at 25 is 6×10^10 . What is the molar solubility of Ba(IO3)2 ?

Question

The solubility product constant (Ksp) of barium iodate, Ba(IO3)2, at 25°C is 6 × 10^10. What is the molar solubility of Ba(IO3)2 in water?

Ocie Christiansen · Accepted Answer

The balanced equation is:Ba(IO3)2 <-> Ba2 + 2IO3I&hellip;.._&hellip;&hellip;&hellip;&hellip;.0&hellip;&hellip;..0C&hellip;_&hellip;&hellip;&hellip;..+x&hellip;&hellip;+2xF&hellip;_&hellip;&hellip;&hellip;&hellip;.x&hellip;&hellip;&hellip;2xOnce balanced, input information at equilibrium:Ksp = [Ba][IO3]^26.0&times;10^-10 = (x)(2x)^26.0&times;10^-10 = 4x^3x = 0.00053 M = 0.00053 mol/LThis value for x also equals the molar solubility (mol/L) of Ba(IO3)2.

Stewart Roberts · Answer

Ksp = [Ba+2][IO3]^2
If x g-ions of Ba+2 are in solution, so are 2x g-ions of the iodate.  Then
 4x^3= 6×10^-10  or x^3=150x^-12.  Then x=5.2×10^-4, which is the molar solubility of the salt.

Anonymous · Answer

Ba Io3 2

Anonymous · Answer

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Ksp = [Ba^2+(aq)][CrO4^2-(aq)] = 2.1*10^-10 [CrO4^2-(aq)] = 2.1*10^-10 / [Ba^2+(aq)] Since BaCrO4 has very low solubility, we can assume that in 0.0015M Na2CrO4(aq), [CrO4^2-(aq)] = 0.0015 mol/litre. So, 1.5*10^-3 = 2.1*10^-10 /[Ba^2+(aq)] then, [Ba^2+(aq)] = 2.1*10^-10 / 1.5*10^-3 = 1.4*10^-7 mol/litre. On dissolving BaCrO4, 1 mole of BaCrO4 produces 1 mole of Ba^2+ ions Therefore, Solubility of BaCrO4 = 1.4*10^-7 mol/litre Since BaCrO4 has Mr = 253 Solubility = 253 * 1.4*10^-7 g/litre = 3.54*10^-5 g/litre.

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The ksp of Ba(IO3)2 at 25 is 6×10^10 . What is the molar solubility of Ba(IO3)2 ?

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