CEXPERIMENT 24: SOLUBILITY PRODUCT CONSTANT Introduction Equilibrium constants f

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CEXPERIMENT 24: SOLUBILITY PRODUCT CONSTANT Introduction Equilibrium constants for the dissociation of sparingly soluble solids are solubility product constants. For the general dissolution reaction of a solid AxBy(s) shown in Reaction (24-1), the solubility product constant is given by Equation (24-2) (24-1) (24-2) It is termed a solubility product constant because it is calculated as the product of the two ionic concentrations (with the appropriate exponents) of the dissociated species. No term for [AxBy] appears in the expression for the solubility product constant because ABy is a pure solid (activity is 1). In this laboratory exercise one of the products of the dissociation of the solid is an anionic weak acid. The compounds used for the exercise have the general formula KHB where B represents the conjugate base of the anionic weak acid. The dissociation that occurs as the solid dissolves is given by Reaction (24-3) and the solubility product constant is shown in Equation (24-4). KHB(s) K.(aq) + HB(a) (24-3) (24-4) In solutions in which the only source of K and HB" is the dissociation shown in Reaction (24- 3), the concentration of K is equivalent to the concentration of HB owing to the ratio of the two products formed in the balanced chemical reaction. By substituting [HB ] for [K'] in Equation (24-4), the Ksp for the salt in pure water solutions becomes Equation (24-5). Ko(in pure water)-[HB-jp IB] = [HB72 (24-5) Since the anion HB is an acid, it's value can be determined by titration with a strong base. The laboratory exercise consists of allowing the solid KHB to reach equilibrium in water, followed by titrating the dissolved HB with standardized NaOH(aq). The net ionic equation for the acid- base titration reaction is shown in Reaction (24-6) HB-(aq) + OH-(aq) H2O( ) + B2-(a) (24-6) After the concentration of HB" is determined from the results of the titration, the solubility product constant is calculated by substituting the concentration into Equation (24-5) A second portion of the laboratory exercise studies the effect of the addition of the common ion K' on the solubility of KHB. In the study a relatively large concentration (0.10 M) of potassium -162-

Explanation / Answer

1)

volume , V = 100.0 mL

= 0.1 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.1*0.1

= 1*10^-2 mol

Molar mass of KNO3 = 1*MM(K) + 1*MM(N) + 3*MM(O)

= 1*39.1 + 1*14.01 + 3*16.0

= 101.11 g/mol

we have below equation to be used:

mass of KNO3,

m = number of mol * molar mass

= 1*10^-2 mol * 101.11 g/mol

= 1.01 g

Answer: 1.01 g

2)

The salt dissolves as:

Mg(OH)2 <----> Mg2+ + 2 OH-

   s 2s

Ksp = [Mg2+][OH-]^2

1.2*10^-11=(s)*(2s)^2

1.2*10^-11= 4(s)^3

s = 1.442*10^-4 M

Molar mass of Mg(OH)2 = 1*MM(Mg) + 2*MM(O) + 2*MM(H)

= 1*24.31 + 2*16.0 + 2*1.008

= 58.326 g/mol

Molar mass of Mg(OH)2= 58.326 g/mol

s = 1.442*10^-4 mol/L

To covert it to g/L, multiply it by molar mass

s = 1.442*10^-4 mol/L * 58.326 g/mol

s = 8.412*10^-3 g/L

Answer:

1.44 mol/L

8.41*10^-3 g/L

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