Question

I need help with the ice table. Please and thank you!

TEEL WILL questions) Give the balanced equation for the solvation of Ca(OH) in water color).(-) e coloq) + 2OH coas + H20lnas Write the solubility-product expression, Ksp, for Ca(OH)2. (cor][ on 3² Ca (OH),ka = [ca(OH)₂] kop= [cator ]² Explain in complete sentences using equations if necessary for illustration at is the stoichiometric relationship of the hydroxide ion concentration to calcium ion concentration? For wery mole of [C ] tad, Slove of our ] are no [OH-] = ² [ca²+] 2 What is the stoichiometric relationship of the solubility of Ca(OH)2 to the concentration of OH in solution? For wery one mole of Coloz .twi mole of cour] will be in added to the solution. 1 [caLOH)₂] = z [our] 3. What is the stoichiometric relationship of the solubility of Ca(OH)2 to the concentration of Ca in solution? For wary wole of [color)], 1 male of Ca 2+ in added to the [uncond ] = [Ca ²+ ] 4. Using LeChatlier's Principle what is the predicted effect of the addition of the common ion, Ca2+, on the solubility of Ca(OH)2? It will deces the solubility of calor) , shifting it to the left. 5. What is the expected effect of the addition of the common ion on Kop? affected by since kap= [Ca 2+ ][or? adding more călciom would mean the T oh would move to decrease in oro to kup the Koo the same. 6. Write the balanced equation for the neutralization of calcium hydroxide by hydrochloric acid. Color), os + 2HCl (ag) - Cach, log) + 2 H20lis

Answer 1

Ans :-

Solubility of Ca(OH)₂ after the addition of 0.2M CaCl₂ :-

Because, CaCl₂ is the strong electrolyte therefore it is completely dissociated unto aqueous solution as :

CaCl₂ (aq) -------------------> Ca²⁺ (aq) + 2 Cl^- (aq)

0.2 M...................................0.2 M..........2x0.2M = 0.4 M

So, concentration of Cl^- = 0.2 M

Let Solubility of Ca(OH)₂ in CaCl₂ = S mol/L

ICE table is :

.........................Ca(OH)₂ (s) <---------------------> Ca²⁺ (aq) ...................+................2 OH^- (aq)

Initial....................Constant.................................0.2 M.............................................0.0 M

Change................Constant..................................+S mol/L.....................................+2S mol/L

Equilibrium .... .....Constant................................(S + 0.2) mol/L..............................2S mol/L

Expression of solubility product (K_sp) is :

K_sp = [Ca²⁺].[OH^-]²

4.09 x 10^-5 M³= (S + 0.2) M. (2S M)²

4.09 x 10^-5 = (S + 0.2) 4S²

4.09 x 10^-5 = 4S³ + 0.8 S²

As, S is very small then neglect 4S³ as it will be so small.

4.09 x 10^-5 = 0.8 S²

S = 7.15 x 10^-3 M

S = 7.15 x 10^-3 mol/1000 mL

S = 7.15 x 10^-4 mol / 100 mL

OR

S = 7.15 x 10^-4 mol x 74.093 g/mol / 100 mL

S = 5.30 x 10^-2 g / 100 mL

Hence, Solubility after the addition of 0.2 CaCl₂ = 5.30 x 10^-2 g /100 mL