Question

The solubility of iron (II) hydroxide, Fe(OH)₂, is 1.43 x10^–3 gram per litre at 25 ⁰C.

(a) Write a balanced equation for the solubility equilibrium.

(b) Write the expression for the solubility product constant, K_sp, and calculate its value.

(c) Calculate the pH of a saturated solution of Fe(OH)₂ at 25 ⁰C.

(d) A 50.0 millilitre sample of 3.00x10^–3 molar FeSO₄ solution is added to 50.0 millilitres of 4.00x10^–6 molar NaOH solution. Does a precipitate of Fe(OH)₂ form? Explain and show calculations to support your answer.

Answer 1

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Dease Student, Given? - solubility of felony = 1643x108g/L at 25°c @ Balanced equation for solubility equilibciom, fe(OH) (s) I feet caq? + 20ni caq? © Kep = [feat] [on] Guven solubility of fe(OH) = 10 43 x 103g1L, to comort it into molarity (mol/L) I mol of fecon = 89.9g = 1.43x1080 x inal IL 89.99 = 1.59x10 mol/L = [fecom [Re2+] = 1.59 x105 mol COH®) 2x(1:59x105) - - 3. 18x15 mark Inus, Kep = (1.59x105m) x (3.18x105m) & = 1.61x1014m3 P.T.O .

③ [OHO ]= 3.18 x 105m. por= -log Cono = -log (3-18x105) = 4.50 pH = 14-por ( pH + por=14) PH = 14 - 4.50 = 9.50 @ when we are mixing too solutions, we will apply, M. x V = M XV (for Pe 24) somex 3.00x103 = Max 100 me [fe²+] = M₂ = some x3.000loin 20. 0015M M, V = 9 X ₂ (for OH) somex 4.00x106 = Mg x loome Lor] = M = some x 4,00 x 106m loome = 2.00x100M for aso, use initial concentrations Osp = [fe²t] [or] P.T.O.

= (0.0015 M) x (2.00x1cm) = 6.00 x 1015 m3 Remember when kep> Qsp (Noppt) and when kep< Osp (ppt fornration) Kep = 1.61x1014 m3 Osp = 6.00 x 100 m² This kep> Qup Hence, no ppt formation Best wishes!!