Question

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

(b) 15.0 mL of 0.292 M Na₂SO₄ and 35.2 mL of 0.200 M KCl

Na⁺
___ M

K⁺
___ M

SO₄^2-
___ M

Cl ^-
___ M


(c) 3.50 g of NaCl in 42.4 mL of 0.586 M CaCl₂ solution
Na⁺
___ M

Ca²⁺
___ M

Cl ^-
___ M

Answer 1

(b) 15.0 mL of 0.292 M Na2SO4 and 35.2 mL of 0.200 M KCl

The reaction:

NOTE that there is no reacton!

therefore, everything is in solution

Find moles of each

Na2SO4 = M1*V1 = 15*0.292= 4.38 mmol

KCl= M2*V2 = 35.2*0.2= 7.04 mmol

note that the final solution = V1+V2 = 15+35.2 = 50.2 ml increase

For each concentrations:

[Na+] = 2*4.38/50.2 = 0.1745

[SO4-2] = 4.38/50.2 = 0.0873

[K+] = 7.04/50.2 = 0.14

[Cl-] = 7.04 /50.2 = 0.14

(c) 3.50 g of NaCl in 42.4 mL of 0.586 M CaCl2 solution

m = 3.5

mol = mass/MW = 3.5/58 = 0.0603 mol of NaCl

M NaCl = mol/V = 0.0603/(42.4/1000) = 1.423 M of NaCl

[Ca+2] = 0.586

[Cl-] = 1.423 + 2*0.586 = 2.59

[NA+] = 1.423