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

Answer:

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

Explanation:

$Moles=Concentration\times Volume (L)$

Concentration of cadmium ion = $[Cd^{2+}]=0.007M$

Volume of cadmium solution = 5 mL = 0.005 L

Moles of cadmium ion = $=0.007 M\times 0.005 L$

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 = $0.008 M\times 0.010 L$

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 :

= $\frac{0.007 M\times 0.005 L}{0.025 L}=0.0014 M$

Initial concentration of thiocyanate ions before reaching equilibrium :

= $\frac{0.008 M\times 0.010 L}{0.025 L}=0.0032 M$