Question

I need help with solving this step by step, please help me solve these step by step and with explanation, please do It on paper and not typed up because it is really confusing when typed up

format and your Solve the following problems. Your solutions should be in the final answers should have the correct units and number of significant figures. 58. proper What mass of Mg(OH), will dissolve in 1.0 L of 0.050 mol/L solution of MgSO4a? (9 marks) a) b) Will a precipitate form when 20.0 mL of Pb(NO3)2(ag) of 2.50 x 10 mol/L is mixed with 45.0 mL of C2CI2(aq) concentration of 1.25 x 10 mol/L? (8 marks) with a concentration with a -3

Answer 1

58.Sol :-

(a).

Molarity = Number of moles of solute / Volume of solution in L

So,

Number of moles of Mg(OH)₂ = Given Molarity of Mg(OH)₂ x Given volume of Mg(OH)₂ solution in L

= 0.050 mol/L x 1.0 L

= 0.050 mol

Also,

Number of moles = Given mass in g / Gram molar mass

So,

Mass of Mg(OH)₂ = Number of moles of Mg(OH)₂ x Gram molar mass of Mg(OH)₂

= 0.050 mol x 58.3197 g/mol

= 2.9 g

Hence, mass of Mg(OH)2 dissolved = 2.9 g

------------------------------

(b).

Pb(NO₃)₂ (aq) + CaCl₂ (aq) -----> PbCl₂ (s) + Ca(NO₃)₂ (aq)

Number of moles of Pb(NO₃)₂ = Molarity x Volume

= 2.50 x 10^-4 mol/L x 0.020 L

= 5.0 x 10^-6 mol

Molarity of Pb(NO₃)₂ after mixing = Moles/Total volume

= 5.0 x 10^-6 mol / (0.020 + 0.045) L

= 7.69 x 10^-5 M

So,

[Pb²⁺] = 1 x [Pb(NO₃)₂] = 7.69 x 10^-5 M

Similarly,

Number of moles of CaCl₂ = Molarity x Volume

= 1.25 x 10^-3 mol/L x 0.045 L

= 5.625 x 10^-5 mol

Molarity of CaCl₂  after mixing = Moles/Total volume

= 5.625 x 10^-5 mol / (0.020 + 0.045) L

= 8.65 x 10^-4 M

So,

[Cl^-] = 2 x [PbCl₂] = 2 x 8.65 x 10^-4 M = 1.73 x 10^-3 M

Precipitates of PbCl₂ are formed after mixing of CaCl₂ and Pb(NO₃)₂ :

Partial dissociation of PbCl₂ in solution is :

PbCl₂ (s) <-----------------------> Pb²⁺ (aq) + 2 Cl^- (aq)

Expression of Reaction quotient (Q_c) (Which is equal to the product of the molar concentration of products raise to power stoichiometric coefficient at any stage of the reaction) is :

Q_c = [Pb²⁺].[Cl^-]²

Q_c = (7.69 x 10^-5 M).(1.73 x 10^-3 M)²

Q_c = 2.3 x 10^-10 M³

Because, Q_c ( 2.3 x 10^-10 M³ ) < K_sp (1.7 x 10^-5 M³)

Therefore, precipitate of PbCl2 are not formed.

NOTE :-

1). If Reaction quotient (Qc) < Solubility product (Ksp) , then precipitate will not takes place 2). If Reaction quotient (Qc) > Solubility product (Ksp) , then precipitate will takes place and 3). If Reaction quotient (Qc) = Solubility product (Ksp) , then reaction is in equilibrium stage that form saturated solution. Solubility product (Ksp) is equal to the product of the molar concentration of products raise to power stoichiometric coefficient at equilibrium stage of the reaction