Problem 1

An unknown salt

\ce{MX2}

is a group 2 metal halide. (a) 10.00 g of

\ce{MX2}

dissolves in 50.0 g of water to give a homogeneous solution whose freezing point is

-4.50\ ^{\circ}\mathrm{C}

. What is the molar mass of

\ce{MX2}

? For water,

K_f = 1.86\ ^{\circ}\mathrm{C}\,m^{-1}

. (b) 10.00 g of

\ce{Na2CO3}

and 10.00 g of

\ce{MX2}

are mixed in 200.0 mL of water and a precipitate of

\ce{MCO3}

forms. What is the pH of the supernatant? The

K_a

\ce{H2CO3}

4.3 \times 10^{-7}

and the

K_a

\ce{HCO3-}

4.7 \times 10^{-11}

. (c) A solution of 10.00 g of

\ce{MX2}

in water is treated with excess silver nitrate; the dried precipitate has mass 15.2 g. What is the identity of

\ce{MX2}

? (d) A sample of 10.00 g of

\ce{MX2}

dissolved in 50 mL of water is treated with increasing amounts of

\ce{Na2SO4}

up to 10 g in total; describe how the mass of precipitate varies with the mass of added

\ce{Na2SO4}

. (e) What colour flame test does

\ce{MX2}

give?

Step 3 of 4: pH of the supernatant

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n(\ce{Na2CO3}) = 0.0943\ \text{mol} > n(\ce{Sr^{2+}}) = 0.0404\ \text{mol} \Rightarrow [\ce{CO3^2-}] = \frac{0.0539}{0.200} = 0.270\ \text{M};\quad \text{pH} \approx 11.9

Analysis

$\ce{SrCO3}$ precipitates quantitatively (Ksp ~ 5.6 × 10 $^{-10}$ ); 0.0943 − 0.0404 = 0.0539 mol of $\ce{CO3^2-}$ remains in 200.0 mL (0.270 M). Hydrolysis $\ce{CO3^2- + H2O <=> HCO3- + OH-}$ with $K_b = K_w/K_{a2} = 2.1 \times 10^{-4}$ gives $[\ce{OH-}] = \sqrt{2.1 \times 10^{-4} \times 0.270} = 7.5 \times 10^{-3}$ M, so pOH ≈ 2.12 and pH ≈ 11.9.