Question

15.) Indicate the concentration of each ion present in the solution formed by mixing the following. (Assume that the volumes are additive.)

(a) 30 mL of 0.100 M HCl and 10.0 mL of 0.420 M HCl

H⁺
M

Cl ^-
M


(b) 15.0 mL of 0.314 M Na₂SO₄ and 23.6 mL of 0.200 M KCl

Na⁺
M

K⁺
M

SO₄^2-
M

Cl ^-
M


(c) 3.50 g of NaCl in 48.4 mL of 0.270 M CaCl₂ solution
Na⁺
M

Ca²⁺
M

Cl ^-
M

Answer 1

In additions the concentraions of ions are given by

M Resultant = [M1V1 + M2V2 ] / (V1 + V2)

a) 30 mL of 0.100 M HCl and 10.0 mL of 0.420 M HCl

H⁺ = [30x 0.1 + 10x0.42]/ (30+10) = 0.18 M

As HCl gives equal H+ and Cl-

[Cl-] in solution = [H+] = 0.18 M

part b)

15.0 mL of 0.314 M Na₂SO₄ and 23.6 mL of 0.200 M KCl
Na2SO4 gives 2 Na+ ions nd one SO4-2 ion

so [Na+] after mixing = 2 x 15 x0.314 / (15+ 23.6) =0.244 M

[K+} = 23.6x0.2 /(15+23.6) =0.122 M

[SO4-2] = 15x0.314 /(15+23.6) =0.122M

[Cl-] = [K+] =0.122 M

part C

3.50 g of NaCl in 48.4 mL of 0.270 M CaCl₂ solution

As there is no solution added the [Ca+2] and [Cl-] remain same as before adding

[Ca+2] = 0.27

[Cl-] = 2 x 0.27 = 0.52 M

molarity of NaCl = moles/V(L)

moles of NaCl = mass/ molar mass = 3.5g/58.5g/mol=0.0598 mol

V = 48.4mL = 0.0484 L

thus molarity of NaCL = 0.0598mol/0.0484 L =1.236 M

Thus [Na+] = [Cl-] = 1.236 M