Question

he solubility product (Ksp) of PbBr2 is 8.9 X 10. Please calculate the molar solubility in: A) Pure water B) 0.20 M Pb(NOs)2 Pb Brs) Pbap +2 Br 8.9- M0T 2 LOZJLES . 45 25 O.Zts Zs 4.45./05 4 4s 0.2x0.2+s 13:105- J S0-60334 9. The solubility of an ionic compound MX (molar mass = 346 g/mol) is 4.63 X 103 g/L. What is the Ksp for this compound? HoW

Answer 1

solubility product of a compound is the mathematical product of its dissolved ion concentration

raised to the power of its stoichiometric coefficients . it is represented by K_sp .

PbBr₂ <==> Pb⁺²(aq) + 2Br^- (aq)

K_sp = [Pb⁺²][Br ^-]²

(a)[Br ^-] = 2[Pb⁺²]

K_sp = [Pb⁺²] X (2[Pb⁺²])²

4[Pb⁺²]³ = 8.9 X 10^-6

[Pb⁺²]³ = 2.225 X 10^-6

Molar solubility  [Pb⁺²] = 0.013 M

(b)  [Pb⁺²] = [Pb(NO₃)₂] = 0.20 M

K_sp = 0.20 X [Br ^-]² =8.9 X 10^-6

[Br ^-] = 6.67 X 10^-3 M

molar solubility = 0.5 X [Br^-]

molar solubility = 0.5 X 6.67 X 10^-3 M

molar solubility = 3.3 X 10^-3 M

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The ionic compound is MX

Solubility of the ionic compound is 4.63 X 10^-3 g/L

molar mass of the compound is 346 g

calculate solubility of the compound in mole per liter as shown below.

Solubility in mol/L = 4.63 X 10^-3 g/L X 1mol MX / 346g

= 1.338 X 10^-5 mol/L

ionization of the compound MX is as follows:

MX M⁺ + X^-

S S S

here, S is the solubility of the compound

Expression of the ionic compound can be written as shown below.

K_sp = [M⁺] [X ^-]

= S X S

= S²

substitute 1.338 X 10^-5 mol /L for S

K_sp= (1.338 X 10^-5 mol /L)²

= 1.79 X 10^-10

therefore the value of the ionic compound is 1.79 X 10^-10

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