Question

Indicate the concentration of each ion or molecule present in the following solutions: (a) 0.25 M NaNO3, ( b ) 1.3 * 10-2 M MgSO4, ( c ) 0.0150 M C6H12O6, (d) a mixture of 45.0 mL of 0.272 M NaCl and 65.0 mL of 0.0247 M 1NH422CO3. Assume that the volumes are additive.

Answer 1

Answer:

For a: The concentration of sodium and nitrate ions are 0.25 M and 0.25 M respectively.

For b: The concentration of magnesium and sulfate ions are $1.3\times 10^{-2}M$ and $1.3\times 10^{-2}M$ respectively.

For c: The concentration of glucose solution is 0.0150 M

For d: The concentration of sodium, chloride, ammonium and carbonate ions in the mixture are 0.111 M, 0.111 M, 0.0146 M and 0.0146 M respectively.

Explanation:

• For a:  0.25 M $NaNO_3$

The chemical equation for the ionization of sodium nitrate follows:

$NaNO_3\rightarrow Na^++NO_3^-$

1 mole of sodium nitrate produces 1 mole of sodium ions and 1 mole of nitrate ions

Concentration of sodium ions = 0.25 M

Concentration of nitrate ions = 0.25 M

Hence, the concentration of sodium and nitrate ions are 0.25 M and 0.25 M respectively.

• For b:  $1.3\times 10^{-2}M$ of $MgSO_4$

The chemical equation for the ionization of magnesium sulfate follows:

$MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}$

1 mole of magnesium sulfate produces 1 mole of magnesium ions and 1 mole of sulfate ions

Concentration of magnesium ions = $1.3\times 10^{-2}M$

Concentration of sulfate ions = $1.3\times 10^{-2}M$

Hence, the concentration of magnesium and sulfate ions are $1.3\times 10^{-2}M$ and $1.3\times 10^{-2}M$ respectively.

• For c:  0.0150 M $C_6H_{12}O_6$

The given compound is a covalent compound and will not dissociate into its respective ions. So, the only chemical species present in the solution is $C_6H_{12}O_6$

Hence, the concentration of glucose solution is 0.0150 M

• For d: A mixture of 45.0 mL of 0.272 M NaCl and 65.0 mL of 0.0247 M $(NH_4)_2CO_3$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$     .....(1)

For sodium chloirde:

Molarity of NaCl solution = 0.272 M

Volume of solution = 45.0 mL

Putting values in equation 1, we get:

$0.272M=\frac{\text{Moles of NaCl}\times 1000}{45}\\\\\text{Moles of NaCl}=\frac{0.272\times 45}{1000}=1.22\times 10^{-2}mol$

For ammonium carbonate:

Molarity of ammonium carbonate solution = 0.0247 M

Volume of solution = 65.0 mL

Putting values in equation 1, we get:

$0.0247M=\frac{\text{Moles of ammonium carbonate}\times 1000}{65}\\\\\text{Moles of ammonium carbonate}=\frac{0.0247\times 65}{1000}=1.61\times 10^{-3}mol$

Total volume of the solution = 45 + 65 = 110 mL

Molarity of NaCl in the mixture = $\frac{1.22\times 10^{-2}mol\times 1000}{110}=0.111M$

Molarity of ammonium carbonate in the mixture = $\frac{1.61\times 10^{-3}mol\times 1000}{110}=0.0146M$

The chemical equation for the ionization of sodium chloride follows:

$NaCl\rightarrow Na^{+}+Cl^{-}$

1 mole of sodium chloride produces 1 mole of sodium ions and 1 mole of chloride ions

Concentration of sodium ions = 0.111 M

Concentration of chloride ions = 0.111 M

The chemical equation for the ionization of ammonium carbonate follows:

$(NH_4)_2CO_3\rightarrow 2NH_4^{+}+CO_3^{2-}$

1 mole of ammonium carbonate produces 1 mole of ammonium ions and 1 mole of carbonate ions

Concentration of ammonium ions = 0.0146 M

Concentration of carbonate ions = 0.0146 M

Hence, the concentration of sodium, chloride, ammonium and carbonate ions in the mixture are 0.111 M, 0.111 M, 0.0146 M and 0.0146 M respectively.