Question

suppose 6.99 g of magnesium acetate (mg(ch₃co₂)₂) are dissolved in 640. ml of water. find the composition of the resulting electrolyte solution.
in particular, list the chemical symbols (including any charge) of each dissolved ion in the table below. list only one ion per row.
then, calculate the concentration of each ion in meq/l, and write the concentration in the second column of each row. be sure you round your answers to the correct number of significant digits.

Explanation:

Step1: Calculate the molar mass of \( Mg(CH_3CO_2)_2 \)

The molar mass of \( Mg = 24.305\space g/mol \), \( C = 12.01\space g/mol \), \( H = 1.008\space g/mol \), \( O = 16.00\space g/mol \).

$$
LATEXBLOCK0
$$

Step2: Calculate the number of moles of \( Mg(CH_3CO_2)_2 \)

$$n=\frac{m}{M}=\frac{6.99\space g}{142.393\space g/mol}\approx0.0491\space mol$$

Step3: Determine the dissociation of \( Mg(CH_3CO_2)_2 \)

\( Mg(CH_3CO_2)_2
ightarrow Mg^{2 +}+2CH_3CO_2^{-} \)
So, \( n(Mg^{2+})=n = 0.0491\space mol \), \( n(CH_3CO_2^{-})=2n=2\times0.0491 = 0.0982\space mol \)

Step4: Calculate the volume of the solution

\( V = 640\space mL=0.640\space L \)

Step5: Calculate the equivalent weight for each ion

For \( Mg^{2+} \), the valence \( z = 2 \), equivalent weight \( E=\frac{M}{z}=\frac{24.305}{2}=12.1525\space g/eq \)
For \( CH_3CO_2^{-} \), the valence \( z = 1 \), equivalent weight \( E=\frac{2\times12.01+3\times1.008 + 2\times16.00}{1}=59.044\space g/eq \)

Step6: Calculate the milliequivalents for each ion

For \( Mg^{2+} \):

$$
LATEXBLOCK1
$$

$$C(Mg^{2+})=\frac{98\space mEq}{0.640\space L}\approx153\space mEq/L$$

For \( CH_3CO_2^{-} \):

$$
LATEXBLOCK2
$$

Answer:

ion	\( \frac{mEq}{L} \)
\( CH_3CO_2^{-} \)	\( 153 \)