Question

onsider the reactin 3E4) The otou in c anda.SMnE.At equslu - - D+4-E 0-0-50. +x+X 3 一112. Since the redox rection;-Cl2 who ya.balone the recha the number o F electrn in the banced half reachon is:? at26°c what is the sorblty Rroduct corstcnt

Answer 1

11. Concentration of substance at equilibrium = Initial concentration - Change in concentration.

Let x be the change in concentration, then

For [C]

0.16 = 0.40 - x

x= 0.40 - 0.16

x = 0.24 M

Now for [E]

Concentration of [E] at equilibrium = Initial concentration - Change in concentration

[E] = 0.50 - x

[E] = 0.50 - 0.24

[E] = 0.26 M

12. Cl₂ goes from oxidation state 0 to oxidation states -1

Cl₂ + 2e- = 2 Cl^-

Cl₂ goes from oxidation state 0 to oxidation states +1

Cl₂ + 2 H₂O = 2 OCl^- + 4 H⁺ + 2e-

Adding above two equations we have

2 Cl₂ + 2 H₂O = 2 Cl^- + 2 OCl^- + 4 H⁺

divide by 2

Cl₂ + H₂O = Cl^- + OCl^- + 2 H⁺

add 2 OH^- on the left and on the right side for charge balancing (remember that 2 H⁺ + 2 OH^- = 2 H₂O)

Cl₂ + 2 OH^- = Cl^- + OCl^- + H₂O

charges, (-2) + (0) => (-1) + (-1) + (0).

So (-2) => (-2). So charge balance is also ok.

Hence no. of electrons are 2

13. Solubility of Mg(OH)₂ in pure water = 0.40 g/L

moles of solute = 0.40 / molar mass

moles of solute =0.40/58 = 0.0068966 = 6.897 x 10^-3 mol

Molarity = mole of solute / Volume in L

Molarity = 6.897 x 10^-3 mol / 1 L = 6.897 x 10^-3 mol/L

K_sp for Mg(OH)₂= [Mg²⁺] x [OH^-]²

Let [Mg²⁺] = [OH^-]² = s

then,

K_sp for Mg(OH)₂ = s x (2s)²= 4s³

K_sp = 4 x (6.897 x 10^-3mol/L)³

K_sp = 13.124 x 10^-7 mol³/L³