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 4 of 4: Sulfate precipitation and flame test

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n(\ce{Sr^{2+}}) = 0.0404\ \text{mol} \Rightarrow m(\ce{SrSO4})_{max} = 0.0404 \times 183.7 = 7.4\ \text{g at } m(\ce{Na2SO4}) = 5.7\ \text{g};\quad \text{flame: crimson red}

Analysis

$\ce{SrSO4}$ (Ksp ≈ 3.4 × 10 $^{-7}$ ) precipitates almost quantitatively: the mass grows nearly linearly with added $\ce{Na2SO4}$ until all Sr $^{2+}$ is consumed at $n(\ce{Na2SO4}) = 0.0404$ mol = 5.7 g — within the 10 g added — then plateaus at ≈ 7.4 g. Strontium colours a flame crimson (scarlet red).