Question

A student working in the laboratory prepared the following reactants: 5 mL of 0.007M Cd2+(aq) 10 mL of 0.008M SCN-(aq) 10 mL of 0.5M HNO3(aq) These reagents were mixed and allowed to stand for 10 minutes. The concentration of Cd(SCN)+ in the resulting equilibrium mixture is found to be 5 x 10−4M. Calculate the initial concentration of Cd2+(aq).

Answer 1

Initial concentration of cadmium ion before reaching equilibrium is 0.0014 M.

Step-by-step explanation:

`Moles=Concentration* Volume (L)`

Concentration of cadmium ion =
`[Cd^(2+)]=0.007M`

Volume of cadmium solution = 5 mL = 0.005 L

Moles of cadmium ion =
`=0.007 M* 0.005 L`

1 mL = 0.001 L

Concentration of thiocyanate ion =
`[SCN^(-)]=0.008 M`

Volume of thiocyanate solution = 10 mL = 0.010 L

Moles of thiocyanate =
`0.008 M* 0.010 L`

Concentration of nitric acid =
`[HNO_3]=0.5 M`

Volume of nitric acid solution = 10 mL = 0.010 L

After mixing all the solution the concentration of cadmium ion and thiocyanate ion will change

Total volume of solution = 0.005 L + 0.010 L + 0.010 L = 0.025 L

Initial concentration of cadmium ion before reaching equilibrium :

=
`(0.007 M* 0.005 L)/(0.025 L)=0.0014 M`

Initial concentration of thiocyanate ions before reaching equilibrium :

=
`(0.008 M* 0.010 L)/(0.025 L)=0.0032 M`