Chemistry Question?
2018-07-07 6:35 am
Determine the concentrations of K2SO4, K , and SO42– in a solution prepared by dissolving 1.52 × 10–4 g K2SO4 in 1.25 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per million (ppm). Note: Determine the formal concentration of SO42–. Ignore any reactions with water.
回答 (2)
2018-07-07 10:23 am
Molar mass of K₂SO₄ = (39.0×2 + 32.1 + 16.0×4) g/mol = 174.1 g/mol
Concentration of K₂SO₄ = [(1.52 × 10⁻⁴ g) / (174.1 g/mol)] / (1.25 L) = 6.98 × 10⁻⁷ M
Each mole of K₂SO₄ contains 2 moles of K⁺ and 1 mole of SO₄²⁻ ions.
Concentration of K⁺ ion = (6.98 × 10⁻⁷ M) × 2 = 14.0 × 10⁻⁶ M
Concentration of SO₄ ion = 6.98 × 10⁻⁷ M
In 1.25 L of the solution :
Mass of K⁺ ion = (1.52 × 10⁻⁴ g) × (39.0×2 / 174.1) = 6.81 × 10⁻⁵ g
Mass of SO₄²⁻ ion = (1.52 × 10⁻⁴ g) × [(32.1 + 16.0×4) / 174.1)] = 8.39 × 10⁻⁵ g
As the solution is very dilute, density of the solution ≈ density of water = 1000 g/L
Mass of 1.25 L of the solution = (1.25 L) × (1000 g/L) = 1250g
Concentration of K⁺ ion in ppm = [(6.81 × 10⁻⁵ g) / (1250 g)] × (1 × 10⁶ ppm) = 0.0545 ppm
Concentration of SO₄²⁻ ion in ppm = [(8.39 × 10⁻⁵ g) / (1250 g)] × (1 × 10⁶ ppm) = 0.0671 ppm
Concentration of K₂SO₄ = [(1.52 × 10⁻⁴ g) / (174.1 g/mol)] / (1.25 L) = 6.98 × 10⁻⁷ M
Each mole of K₂SO₄ contains 2 moles of K⁺ and 1 mole of SO₄²⁻ ions.
Concentration of K⁺ ion = (6.98 × 10⁻⁷ M) × 2 = 14.0 × 10⁻⁶ M
Concentration of SO₄ ion = 6.98 × 10⁻⁷ M
In 1.25 L of the solution :
Mass of K⁺ ion = (1.52 × 10⁻⁴ g) × (39.0×2 / 174.1) = 6.81 × 10⁻⁵ g
Mass of SO₄²⁻ ion = (1.52 × 10⁻⁴ g) × [(32.1 + 16.0×4) / 174.1)] = 8.39 × 10⁻⁵ g
As the solution is very dilute, density of the solution ≈ density of water = 1000 g/L
Mass of 1.25 L of the solution = (1.25 L) × (1000 g/L) = 1250g
Concentration of K⁺ ion in ppm = [(6.81 × 10⁻⁵ g) / (1250 g)] × (1 × 10⁶ ppm) = 0.0545 ppm
Concentration of SO₄²⁻ ion in ppm = [(8.39 × 10⁻⁵ g) / (1250 g)] × (1 × 10⁶ ppm) = 0.0671 ppm
2018-07-07 10:04 am
Determine the concentration.....
1.52x10^-4g K2SO4 x (1 mol K2SO4 / 174.3g K2SO4) / 1.25L = 6.98x10^-7 mol K2SO4 / 1L
Since K2SO4 ionizes completely to form K+ ions and SO4^2- ions and there is no appreciable hydrolysis of SO4^2- ions, then....
[K2SO4] = 6.98x10^-7M
[K+] = 2 x 9.0x10^-7 = 1.40x10^-6M
[SO4^2-] = 6.98x10^-7M
Parts per million is the same as milligrams per liter.
1.52x10^-4g x (1000 mg / 1g) / 1.25L = 0.122 mg/L .... or .... 0.122 ppm
Parts per million is the mass of the solute per the mass of solution, times 1 million.
1.52x10^-4g / 1250g x (1x10^6) = 0.122 ppm
Find the mass of K in 1.52x10^-4g K2SO4
1.52x10^-4g K2SO4 x (78.2g K / 174.3g K2SO4) / 1250g x (1x10^6) = 5.46x10^-2 ppm
Find the mass of SO4 in 1.52x10^-4g K2SO4
1.52x10^-4g K2SO4 x (96.1g SO4 / 174.3g K2SO4) / 1250g x (1x10^6) = 6.70x10^-2 ppm
1.52x10^-4g K2SO4 x (1 mol K2SO4 / 174.3g K2SO4) / 1.25L = 6.98x10^-7 mol K2SO4 / 1L
Since K2SO4 ionizes completely to form K+ ions and SO4^2- ions and there is no appreciable hydrolysis of SO4^2- ions, then....
[K2SO4] = 6.98x10^-7M
[K+] = 2 x 9.0x10^-7 = 1.40x10^-6M
[SO4^2-] = 6.98x10^-7M
Parts per million is the same as milligrams per liter.
1.52x10^-4g x (1000 mg / 1g) / 1.25L = 0.122 mg/L .... or .... 0.122 ppm
Parts per million is the mass of the solute per the mass of solution, times 1 million.
1.52x10^-4g / 1250g x (1x10^6) = 0.122 ppm
Find the mass of K in 1.52x10^-4g K2SO4
1.52x10^-4g K2SO4 x (78.2g K / 174.3g K2SO4) / 1250g x (1x10^6) = 5.46x10^-2 ppm
Find the mass of SO4 in 1.52x10^-4g K2SO4
1.52x10^-4g K2SO4 x (96.1g SO4 / 174.3g K2SO4) / 1250g x (1x10^6) = 6.70x10^-2 ppm
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