Calculate the mass (g) of potassium nitrate KNO_3 that is required to prepare 10

Question

Calculate the mass (g) of potassium nitrate KNO_3 that is required to prepare 100.0 mL of 0.10 um M KNO_3 (aq). The solubility product constant for Mg(OH)_2(s) is 1.2 times 10^-11 Calculate the solubility of magnesium hydroxide in molL and in g/L. The concentration of Pb^2+ ions and of CO^2+ _3 ions in a saturated aqueous solution of PbCO_3 is 1.8 times 10^-7 M. (a) Determine K_sp of PbCO_3. (b) Calculate its solubility in g/L and in g 100 mL. Determine the molar solubility of AgCl in (a) pure water and (b) 0.10 M NaCl(aq). The K_sp of AgCI is 1.8 times 10^-10.

Explanation / Answer

a.) AgCl dissociates in water as:

AgCl -------> Ag+ + Cl-

Ksp = [Ag+] [Cl-]

1.8 X 10-10 = [s] [s]

1.8 X 10-10 = s2

s = 1.342 X 10-5 M or s = 1.342 X 10-5 mol/L

b.) We are now dissolving AgCl in 0.10 M NaCl(aq). It instantly tells us that Cl- is the common ion between the two.

NaCl completely dissociates in water to form Na+ and Cl- ions.

NaCl --------> Na+ + Cl-

Concentration of the both Na+ and Cl- ions in the aqueous solution is 0.10 M as given in the question.

Now we add AgCl to the aqueous solution of NaCl.

The concentration of Ag+ ions will be 'x' molar at equilibrium. The concentration of Cl- ions will be '0.10 + x' as 0.10 M Cl- ions were already present in the solution thanks to NaCl.

So, the Ksp equation for AgCl can be written as:

Ksp = [Ag+] [Cl-]

Ksp = [x] [0.10 +x]

As we know from common ion effect, the solubility of the ion decreases when a common ion is present. Here the common ion is Cl-. So, the '0.10+x' concentration of Cl- reduces to '0.10' as 'x' is too small to be taken into account.

1.8 X 10-10 = [x] [0.10]

x = 1.8 X 10-9 M

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