Question

Indicate the concentration of each ion present in the solution formed by mixing the following.

A. 42.0 mL of 0.150 M NaOH and 37.6 mL of 0.410 M NaOH. Assume that the volumes are additive.

B.  44.0 mL of 0.140 M Na2SO4 and 25.0 mL of 0.160 M KCl. Assume that the volumes are additive.

C.  3.20 g KCl in 75.0 mL of 0.220 M CaCl2 solution. Assume that the volumes are additive.

Answer 1

A.Number of moles of NaOH in first solution = M*V = 0.150 M * 42.0 mL = 6.3 mmol

Number of moles of NaOH in 2nd solution = M*V = 0.410 M * 37.6 mL = 15.416 mmol

Total volume of the solution = 42.0 + 37.6 = 79.6 mL

Total number of moles of NaOH = 6.3 + 15.416 = 21.716 mmol

Net conc. of NaOH = moles/volume = 21.716 mmol/79.6 mL = 0.272814 M = 0.273 M

[Na^+] = 0.273 M

[OH^–] = 0.273 M

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B. Number of moles of Na₂SO₄ = M*V = 0.140 M * 44.0 mL = 6.16 mmol

Number of moles of KCl = M*V = 0.160 M * 25.0 mL = 4 mmol

Total volume of the solution = 44.0 + 25.0 = 69 mL

Net conc. of Na₂SO₄ = moles/volume = 6.16 mmol/ 69 mL = 0.089275 M

[Na⁺] = 2 * 0.089275 = 0.178551 M = 0.179 M

[SO₄^–2] = 0.089275 M = 0.0893 M

Net conc. of KCl = moles/volume = 4 mmol/ 69 mL = 0.057971 M

[K⁺] = 0.057971 M = 0.0580 M

[Cl^–] = 0.0580 M

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C. Number of moles of KCl = mass/mol.wt.= 3.20 g/(74.55 g/mol) = 0.042924 mol = 42.92421 mmol

Number of moles of CaCl₂ = M*V = 0.220 M * 75.0 mL = 16.5 mmol

Total volume of the solution = 75.0 mL

Net conc. of KCl = moles/volume = 42.92421 mmol/ 75.0 mL = 0.572323 M

[K⁺] = 0.572323 M = 0.572 M

[Cl^–] = 0.572 M

Net conc. of CaCl₂ = moles/volume = 16.5 mmol/ 75.0 mL = 0.220 M

[Ca⁺²] = 0.220 M

[Cl^–] = 2* 0.220 M = 0.440 M

[K+] = 0.572 M

[Ca⁺²] = 0.220 M

[Cl^–] = 0.572 + 0.440 = 1.012 M