Question

Q: 50.00 ml of 0.5216 M copper(II) nitrate solution is combined with 100.0 ml of 0.5580 M potassium hydroxide solution according to the following unbalanced equation.

Cu(NO3)2 (ag) + KOH -> Cu(OH)2 (s) + KNO3 (aq)

(a) How many grams of copper(II) hydroxide can be formed?

(b) The solid was filtered, dried, and found to have a mass of 2.420 g. What was the percent yield?

(c) What are the concentrations of all species in the solution before the solid is removed?

Answer 1

Cu(NO3)2 (ag) + KOH -> Cu(OH)2 (s) + KNO3 (aq)

Balance equation

Cu(NO3)2 (ag) + 2KOH -> Cu(OH)2 (s) + 2KNO3 (aq)

(a) How many grams of copper(II) hydroxide can be formed?

find moles reacting, limitng reactant, and then moles of Cu(OH)2 produced

mol of Cu(NO3)2 = M*V = 50*0.5216 = 26.08 mmol of Copper Nitrate

mol of KOH = M*V = 100*0.558 = 55.8 mmol of base

2 mol of base : 1 mol of Copper Nitrate

the limiting reactants is the copper nitrate

26.08 mmol of Copper Nitrate need 2X that of base

2*26.08 = 52.16 mmol of Base

All Copper Nitrate will react.

1 mol of Cu(OH)2 per 1 mol of Cu(NO3)

therefore, expect:

26.08 mmol of Cu(OH)2 to form

but we need grams so MW of Cu(NO3)2

MW of Cu(NO3)2 = 97.561

mass = mol*MW = (26.08*10^-3) * 97.561 = 2.544 grams of Cu(OH)2 will be produced

b)

IF real value of Cu(OH)2 = 2.42g then

% yield = real/theoreteical * 100% = 2.42/2.544 *100 = 95.11%

c)

Concnetration of all species:

M = mol/V

VT = V1+V2 = 50 ml + 100 ml = 150 ml

calculate for each

Cu(NO3)2 --> 0 M since all reacted

KOH --> Initial - reacted = final

55.8 -52.16 = 3.64 mmol of KOH left

M = mmol / ml = 3.64/150 = 0.02426 M of KOH

Cu(OH)2 --> 0M since in theory, everything is solid

For

KNO3 (aq) --> 52.16 mmol reacted

M = 52.16 mmol / 150 ml = 0.3477 mol of KNO3 per liter