Question

Be sure to answer all parts.

If 2.25 g of CuSO₄ is dissolved in 9.79 × 10² mL of 0.380 M NH₃, calculate the concentrations of the following species at equilibrium.

Cu²⁺

___× 10^___ M (Enter your answer in scientific notation.)

NH₃

___ M

Cu(NH₃)₄²⁺

___ M

Answer 1

Moles of CuSO4 = amount in g/ molar mass

= 2.25 g/159.609 g/mol

= 0.014 mol

One mole of CuSO4 = Cu2+ + SO42-

0.014 mol CuSO4 0.014 mol Cu2+

Molarity = moles / volume in L

9.79 × 10² mL = 0.979 L

Cu2+:0.014 mol Cu2+/ 0.979 L

=0.014 M or 1.4*10^-2 M

NH3 0.380 M

Here Cu 2+ is limiting agent.

The balance equation :
CuSO4 + 4NH3 --> Cu(NH3)4^2+ + SO4-

0.014 mol Cu2+ * Cu(NH3)4^2+ / 1 mol Cu2+

= 0.014 mol Cu(NH3)4^2+

Molarity = moles / volume in L

9.79 × 10² mL = 0.979 L

Cu(NH3)4^2+

:0.014 mol Cu(NH3)4^2+/ 0.979 L

=0.014 M or 1.4*10^-2 M Cu(NH3)4^2+