Question

Suppose 0.250 g of cobalt(II) chloride hexahydrate is dissolved in 50.0 mL of 4 M HCl, and heated to 80^oC in a cuvette having a 1.00-cm path length. At this temperature, the percent transmittance of the solution at 690 nm is measured to be 48.0%.

What is the equilibrium concentration of CoCl₄^2- in the solution? (Hint: use Beer's Law)

a. 2.10 × 10^-2 M

b. 8.25 × 10^-2 M

c. 1.81 × 10³ M

d. 5.52 × 10^-4 M

What is the equilibrium concentration of Co(H₂O)₆²⁺?

a. 5.52 × 10^-4 M

b. 2.16 × 10^-2 M

c. 2.05 × 10^-2 M

d. 2.11 × 10^-2 M

Answer 1

1st part

Absorbance, A = 2 - log₁₀ %T = 2 - log₁₀ 48 = 0.32

Again A=ebc

e is the molar absorbtivity with units of L mol^-1 cm^{-1
b is the path length of the sample andc is the concentration of the compound in solution, expressed in mol L-1}

Therefore c = A/eb = 0.32/ e x 1 = 0.32/ e = 0.32/ 557.2 = 5.54 x 10-4 M

[The reported molar absorptivity value for [CoCl4]2- at around 700 nm is 577.2 M-1cm-1. ]

So d is the answer

2nd part:

Conc. of Co(H2O)6 2+ = (conc. of cobalt(II) chloride hexahydrate) - (conc. of [CoCl₄]^2-)

= (0.250/ 237.93)/(50/1000) - 5.54 x 10-4 M = 2.05 x 10^-2 M so C is the answer