Question

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

(a) 40 mL of 0.100 M HCl and 10.0 mL of 0.600 M HCl

H⁺ ________ M

Cl ^- ________ M


(b) 15.0 mL of 0.300 M Na₂SO₄ and 20.6 mL of 0.200 M KCl

Na⁺ _________ M

K⁺ __________ M

SO₄^2- __________ M

Cl ^- __________ M


(c) 3.50 g of NaCl in 40.6 mL of 0.365 M CaCl₂ solution
Na⁺ ___________ M

Ca²⁺ __________ M

Cl ^- __________ M

Answer 1

(a) 40 mL of 0.100 M HCl and 10.0 mL of 0.600 M HCl
Number of moles of 40 mL of 0.100 M HCl , n = molarity x volume in L

                                                                  = 0.100 M x 0.040 L

                                                                  = 0.004 moles

Number of moles of 10.0 mL of 0.600 M HCl, n' = molarity x volume in L

                                                                  = 0.600 M x 0.010 L

                                                                  = 0.006 moles

total number of moles , N = n+ n' = 0.004+0.006 = 0.010 moles

Total volume of the mixture , V = 40.0+10.0 = 50.0 mL = 0.050 L

So concentration of HCl in the mixture is M = N / V

                                                               = 0.010 mol / 0.050 L

                                                               = 0.2 M

1 mole HCl contains 1 mole H⁺ & 1 mole Cl^-

So [H⁺] = [Cl^-] = [HCl] = 0.2M

(b) 15.0 mL of 0.300 M Na₂SO₄ and 20.6 mL of 0.200 M KCl

Number of moles of Na₂SO₄ is = molarity x volume in L = 0.300 M x 0.015 L = 0.0045 mol

Number of moles of KCl is = Molarity x volume in L = 0.200 M x 0.0206 L = 0.00412 mol

Total volume of the mixture = 15.0 mL + 20.6 mL = 35.6 mL = 0.0356 L

So concentration of Na₂SO₄ in the mixture = number moles / total volume

                                                              = 0.0045 mol / 0.0356 L

                                                              = 0.126 M

1 mole of Na₂SO₄ contains 2 moles of Na⁺ & 1 mole of SO₄^2-

So [ Na⁺ ] = 2x[Na₂SO₄] = 2 x0.126 = 0.252 M

   [ SO₄^2- ] = [Na₂SO₄] = 0.126 M

Concentration of KCl in the mixture = number of moles / total volume

                                                   = 0.00412 mol / 0.0356 L

                                                   = 0.116 M

1 mole of KCl has 1 mole of K⁺ & 1 mole of Cl^-

[K⁺] = [Cl^-] = [KCl] = 0.116 M

(c) 3.50 g of NaCl in 40.6 mL of 0.365 M CaCl₂ solution

Number of moles of NaCl = mass / molar mass

                                     = 3.50 g / 58.5 (g/mol)

                                     = 0.06 mol

Number of moles of CaCl₂ is = Molarity xvolume in L

                                          = 0.365 M x 0.0406 L

                                          = 0.015 mol

Total number of moles of Cl^- , n = n_Cl- (from NaCl) + nCl^- (from CaCl₂)

                                             = 0.06 + (2x0.015 )               Since 1 mole of CaCl₂ contains 2 moles of Cl^-

                                                = 0.09 mol

Total volume of the mixture,V = 40.6 mL = 0.0406 L

So [Cl^-] = n / V

            = 0.09 mol / 0.0406 L

            = 2.22 M

[Na⁺] = [NaCl] ( since 1 mole of NaCl contains 1 mole of Na⁺)

        = 0.06 mol / 0.0406 L

        = 1.48 M

[Ca²⁺] = [CaCl₂ ] = 0.015 mol / 0.0406 L

                            = 0.37 M