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 2 of 4: Halide identity from AgX mass

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\frac{10.00}{M_M + 2M_X} \times 2 (107.87 + M_X) = 15.2 \Rightarrow M_X = 79.9\ (\ce{Br}),\ M_M = 87.6\ (\ce{Sr}) \Rightarrow \ce{MX2} = \ce{SrBr2}

Analysis

Each mole of MX2 gives 2 moles of AgX. Testing the halides against the molar mass ~248 from part (a): with X = Br, $M(\ce{SrBr2}) = 247.4$ g mol $^{-1}$ , so $n(\ce{SrBr2}) = 10.00/247.4 = 0.0404$ mol, $n(\ce{AgBr}) = 0.0808$ mol and $m(\ce{AgBr}) = 0.0808 \times 187.8 = 15.2$ g — exact match. The salt is strontium bromide, $\ce{SrBr2}$ .

Common pitfall. Do not forget the stoichiometric factor 2:

n(\ce{AgX}) = 2n(\ce{MX2})

; omitting it doubles the inferred mass of X.