Question

At a certain temperature, the concentration of I2 in its saturated aqueous solution is 1.300×10−3 M, and the equilibrium achieved when I2 distributes itself between H2O and CCl4 is I2(aq)⇌I2(CCl4),K=85.5.A 13.0-mL sample of saturated I2(aq) is shaken with 13.0 mL of CCl4. After equilibrium is established, the two liquid layers are separated. How many milligrams of I2 will be in the aqueous layer? Express your answer to two significant figures and include the appropriate units.

Answer 1

K = 85.5 = [I1] aq / [I2] CCl4

Initial I2 conc = 0.0013 M , volume of sample = 13 ml = 0.013 L

Moles of I2 = M x V = 0.0013 x 0.013 = 1.69 x 10^ -5

Let m be moles of I2 in CCL4 , then moles of I2 left in aquoes = 1.69 x 10^ -5 - m

[I2] in aq = Moles/vol of aq in L = ( 1.69x10^ -5 -m) /0.013

[I2] CCl4 = ( m/0.013)

K = 85.5 = ( 1.69x10^-5 -m) /0.013) / ( m/0.013)

85.5 = ( 1.69x10^-5 -m) / m

85.5 m = 1.69x10^-5 -m

m = 1.9537 x 10^ -7

I2 mass = moles x molar mass of I2 = 1.9537 x10^ -7 x 126.9 = 2.48 x 10^ -5 g = 2.48 x 10^ -2 mg = 0.0248 mg

      = 0.025 mg