Question

Calculate the mass (in mg) of solid chromium(III) nitrate needed to make 250. mL of an aqueous solution with a pH of 4.00. The Ka of Cr(H2O)63+ is 1.0 x 10−4.

48 mg

3.0 mg

12 mg

6.0 mg

24 mg

Answer 1

we need an aqueous solution of [Cr(H2O)₆]³⁺ with pH of 4.00

pH = - log[H3O⁺] = 4.00

=> [H3O⁺] = 10^-4.00

Let the concentration of [Cr(H2O)₆]³⁺ giving pH 4 be 'C' M

The dissociation reaction of [Cr(H2O)₆]³⁺ is

---------------- [Cr(H2O)₆]³⁺ + H2O ----- >  [Cr(OH)(H2O)₅]³⁺ + H3O⁺ , Ka = 1.0x10^-4

Initial conc(M): C ------------------------------ 0, ------------------------ 0

Eqm.Conc(M):(C - y) ,------------------------ y, ------------------------- y

Given  [H3O⁺] = y = 10^-4.00

Also Ka = 1.0x10^-4 = [[Cr(OH)(H2O)₅]³⁺ ]x [H3O⁺] / [[Cr(H2O)₆]³⁺] = yxy / (C - y)

=> 1.0x10^-4 = (10^-4.00 )² / (C - 10^-4.00 )

=> (C - 10^-4.00 ) = (10^-4.00 )² / 1.0x10^-4 = 10^-4.00

=> C = 2 x 10^-4.00 = 2x10^-4 M

Given the volume of the aqueous solution of  [Cr(H2O)₆]³⁺ = 250 mL = 0.250L

Hence moles of  [Cr(H2O)₆]³⁺ = CxV(L) = 2x10^-4 M x  0.250L = 5x10^-5 mol

Since 1 molecule of [Cr(H2O)₆]³⁺ contains 1 Cr³⁺ ion, to produce  5x10^-5 mol of [Cr(H2O)₆]³⁺ we will also need  5x10^-5 mol of chromium(III) nitrate [Cr(NO3)3].

Hence moles of chromium(III) nitrate [Cr(NO3)3] needed =  5x10^-5 mol

Molecular mass of  chromium(III) nitrate [Cr(NO3)3] = 238.0 g/mol

Hence mass of  chromium(III) nitrate [Cr(NO3)3] needed = 5x10^-5 mol x238.0 g/mol = 1.19 x10^-2 g

= 1.19 x10^-2 g x (1000 mg / 1g) = 11.9 mg = 12 mg (answer)

Hence 3rd option is correct