Question

100 mL of .220 M copper ll chloride is mixed with 100 mL of .750 M sodium hydroxide according to the following equation:

CuCl2 + 2NaOH -->Cu(OH)2 + 2NaCl

1. What is the mass of solid product formed?

2. Calculate the concentration of Cu2+ remaining in solution.

3. Calculate the concentration of OH- remaining in solution.

4. Calculate the concentration of Cl- remaining in solution

5. Calculate the concentration of Na+ remaining in solution.

Answer 1

CuCl2 + 2NaOH -->Cu(OH)2 + 2NaCl

100 mL of 0.220 M CuCl2 = 100 mL * ( 0.220 mol /1000 mL) => 0.022 mol

100 mL of 0.750 M NaOH = 100 mL * ( 0.750 mol / 1000 mL) => 0.075 mol

1 eq of CuCl2 requires 2 eq of NaOH

So 0.022 mol will require 2*0.022 mol = 0.044 mol of NaOH

1 eq of CuCl2 produces 1 eq of Cu(OH)2

1)

0.022 mol CuCl2 produces 0.022 mol of Cu(OH)2

molar mass of Cu(OH)2 = 97.56 g/mol

0.022 mol of Cu(OH)2 = 0.022 mol * 97.56 g/mol => 2.15 g of solid product would be formed.

2) concentration of Cu2+ remaining in solution

All the CuCl2 completely reacted with excess NaOH .

So there will be no Cu2+ ions presents in solution.

3) Calculate the concentration of OH- remaining in solution

Total moles of OH- = 0.075

Reacted moles of OH- = 0.044

Remaining moles of OH- = 0.075 - 0.044 => 0.031

Total volume of solution 200 mL

Concentration of [OH-] = 0.031 mol/0.200 L => 0.155 M

4) concentration of Cl- remaining in solution

0.044 mol NaOH will produce 0.044 mol of NaCl

concentration of Cl- = 0.044 mol / 0.200 Ltr => 0.22 M

5) concentration of Na+ remaining in solution.

Na+ ions in both form NaOH and NaCl exists as Na+ ions

so 0.075 mol / 0.200 Ltr => 0.375 M