Respuesta :
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 [tex]1.3\times 10^{-2}M[/tex] and [tex]1.3\times 10^{-2}M[/tex] 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 [tex]NaNO_3[/tex]
The chemical equation for the ionization of sodium nitrate follows:
[tex]NaNO_3\rightarrow Na^++NO_3^-[/tex]
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: [tex]1.3\times 10^{-2}M[/tex] of [tex]MgSO_4[/tex]
The chemical equation for the ionization of magnesium sulfate follows:
[tex]MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}[/tex]
1 mole of magnesium sulfate produces 1 mole of magnesium ions and 1 mole of sulfate ions
Concentration of magnesium ions = [tex]1.3\times 10^{-2}M[/tex]
Concentration of sulfate ions = [tex]1.3\times 10^{-2}M[/tex]
Hence, the concentration of magnesium and sulfate ions are [tex]1.3\times 10^{-2}M[/tex] and [tex]1.3\times 10^{-2}M[/tex] respectively.
- For c: 0.0150 M [tex]C_6H_{12}O_6[/tex]
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 [tex]C_6H_{12}O_6[/tex]
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 [tex](NH_4)_2CO_3[/tex]
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex] .....(1)
For sodium chloirde:
Molarity of NaCl solution = 0.272 M
Volume of solution = 45.0 mL
Putting values in equation 1, we get:
[tex]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[/tex]
For ammonium carbonate:
Molarity of ammonium carbonate solution = 0.0247 M
Volume of solution = 65.0 mL
Putting values in equation 1, we get:
[tex]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[/tex]
Total volume of the solution = 45 + 65 = 110 mL
Molarity of NaCl in the mixture = [tex]\frac{1.22\times 10^{-2}mol\times 1000}{110}=0.111M[/tex]
Molarity of ammonium carbonate in the mixture = [tex]\frac{1.61\times 10^{-3}mol\times 1000}{110}=0.0146M[/tex]
The chemical equation for the ionization of sodium chloride follows:
[tex]NaCl\rightarrow Na^{+}+Cl^{-}[/tex]
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:
[tex](NH_4)_2CO_3\rightarrow 2NH_4^{+}+CO_3^{2-}[/tex]
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.
A) The concentration of each ion is;
Concentration of Na⁺ = 0.25 M
Concentration of NO₃⁻ = 0.25 M
B) The concentration of each ion is;
Concentration of Mg²⁺ = 1.3 × 10⁻² M
Concentration of SO₄²⁻ = 1.3 × 10⁻² M
C) Concentration remains 0.015 M.
D) The concentration of each ion is;
Concentration of Na⁺ = 0.111 M
Concentration of Cl⁻ = 0.111 M
Concentration of NH₄⁺ = 0.0146 M
Concentration of CO₃⁻² = 0.0292 M
A) We are given the solution NaNO₃.
Its' molarity = 0.25 M
To know the concentration of each ion or molecule present, we have to write its' ionization equation first;
NaNO₃ → Na⁺ + NO₃⁻
Now, the equation is balanced and we can see that;
1 mole of NaNO₃ produces 1 mole of Na⁺ and 1 mole of NO₃⁻.
Thus;
Concentration of Na⁺ = 0.25 M
Concentration of NO₃⁻ = 0.25 M
B) We are given the solution MgSO₄
Its' molarity = 1.3 × 10⁻² M
It's ionization equation is;
MgSO₄ → Mg²⁺ + SO₄²⁻
The equation is balanced and so we can say that;
1 mole of MgSO₄ produces 1 mole of Mg²⁺ and 1 mole of SO₄²⁻.
Thus;
Concentration of Mg²⁺ = 1.3 × 10⁻² M
Concentration of SO₄²⁻ = 1.3 × 10⁻² M
C) We are given the solution C₆H₁₂O₆
Its' molarity = 0.015 M
This solution is called glucose and it remains intact when put in water and as a result, it has ionization of 1.
Thus, the concentration remains 0.015 M.
D) We are given a mixture of;
45 mL of 0.272M NaCl and 65.0 mL of 0.0247M (NH₄)₂CO₃.
Let us first find the number of moles for each substance using the formula;
Number of moles = (molarity × volume)/1000
For NaCl; Number of moles = (0.272 × 45)/1000 = 0.01224 mol
For (NH₄)₂CO₃; Number of moles = (0.0247 × 65)/1000 = 0.0016055 mol
Total volume of solution = 45 + 65 = 100 mL
Molarity of each substance in the solution is gotten from the formula;
Molarity = (Number of moles × 1000)/Total volume
Molarity of NaCl = (0.01224 × 1000)/110 = 0.111 M
Molarity of (NH₄)₂CO₃ = (0.0016055 × 1000)/110 = 0.0146 M
The ionization of NaCl gives;
NaCl → Na⁺ + Cl⁻
The equation is balanced and so we can say that;
1 mole of NaCl produces 1 mole of Na⁺ and 1 mole of Cl⁻.
Thus;
Concentration of Na⁺ = 0.111 M
Concentration of Cl⁻ = 0.111 M
The ionization of NaCl gives;
(NH₄)₂CO₃ → 2NH₄⁺ + CO₃⁻²
The equation is balanced and so we can say that;
1 mole of (NH₄)₂CO₃ produces 2 mole of NH₄⁺ and 1 mole of CO₃⁻².
Thus;
Concentration of NH₄⁺ = 0.0146 M
Concentration of CO₃⁻² = 2 × 0.0146 M = 0.0292 M
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