Answer:
= 0.014 g of BaCO3
Explanation:
Let x = mol/L of BaCO3 that dissolve.
This will give;
x mol/L Ba2+ and x mol/L CO32-
But;
Ksp = 5.1x10^-9.
Therefore;
Ksp = 5.1 x 10^-9 = (x)(x)
Thus;
x = molar solubility
= √ (5.1 x 10^-9)
= 7.1 x 10^-5 M
Therefore;
Mass BaCO3 = 7.1 x 10^-5 M x 1 L x 197.34 g/mol
= 0.014 g
The mass of BaCO3 that dissolves in 1000 ml of water is 0.014 g.
The term Ksp shows the extent of dissolution of a substance in solution. In this case we are told in the question that the Ksp of the BaCO3 is 5.1x10^-9. Recall that the mass that dissolve is 1000 mL of water is the solubility of the solute.
Ksp = [Ba^2+] [CO3^2-] = x^2
x =√ Ksp
x = √5.1x10^-9
x = 7.14 * 10^-5 M
moles = concentration * volume
Mass/molar mass = concentration * volume
Mass = concentration * volume * molar mass
Mass = 7.14 * 10^-5 M * 1 L * 197 g/mol
Mass = 0.014 g
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