## Solution Stoichiometry

58) (a) [HCl]1 = 0.100 M          [HCl]2 = 0.500 M

VHCl1 = 20.0 mL           VHCl2 = 10.0 mL

HCl(aq) H+(aq) + Cl-(aq)

[HCl]1 = n1/VHCl1

n1 = [HCl]1 x VHCl1 = 0.100 mol HCl/L x 20.0 mL x 1 L/103 mL = 2.00 x 10-3 mol HCl

n2 = [HCl]2 x VHCl2 = 0.500 mol HCl/L x 10.0 mL x 1 L/103 mL = 5.00 x 10-3 mol HCl

[HCl] = nT/VT = (2.00 x 10-3 mol HCl + 5.00 x 10-3 mol HCl)/((20.0 mL + 10.0 mL) x 1 L/103 mL) = 0.233 M

[H+] = 0.233 mol HCl/L x 1 mol H+/1 mol HCl = 0.233 M

[Cl -] = 0.233 mol HCl/L x 1 mol Cl -/1 mol HCl = 0.233 M

(b) [Na2SO4] = 0.300 M          [KCl] = 0.200 M

VNa2SO4 = 15.0 mL            VKCl = 10.0 mL

Na2SO4(aq) 2Na+(aq) + SO42-(aq)

KCl(aq) K+(aq) + Cl-(aq)

[Na2SO4] = n/V

nNa2SO4 = [Na2SO4] x V = 0.300 mol Na2SO4/L x 15.0 mL x 1 L/103 mL = 4.50 x 10-3 mol Na2SO4

nKCl = [KCl] x V = 0.200 mol KCl/L x 10.0 mL x 1 L/103 mL = 2.00 x 10-3 mol KCl

[Na2SO4] = nT/VT = 4.50 x 10-3 mol Na2SO4/(25.0 mL x 1 L/103 mL) = 0.180 M Na2SO4

[KCl] = 2.00 x 10-3 mol KCl/(25.0 mL x 1 L/103 mL) = 0.0800 M KCl

[Na+] = 0.180 mol Na2SO4/L x 2 mol Na+/1 mol Na2SO4 = 0.360 M

[SO42-] = 0.180 mol Na2SO4/L x 1 mol SO42-/1 mol Na2SO4 = 0.180 M

[K+] = 0.0800 mol KCl /L x 1 mol K+/1 mol KCl = 0.0800 M

[Cl-] = 0.0800 mol KCl /L x 1 mol Cl-/1 mol KCl = 0.0800 M

(c) m = 3.50 g NaCl          [CaCl2] = 0.500 M          V = 50.0 mL

NaCl(aq) Na+(aq) + Cl-(aq)

CaCl2(aq) Ca2+(aq) + 2Cl-(aq)

nNaCl = 3.50 g NaCl x 1 mol NaCl/58.44 g NaCl = 0.0599 mol NaCl

nCaCl2  = 0.500 mol CaCl2/L x 50.0 mL x 1 L/103 mL = 0.0250 mol CaCl2

[Na+] = n/VT = 0.0599 mol NaCl/(50.0 ml x 1 L/103 mL) = 1.20 M

[Cl-] = (0.0599 mol Cl- + 2 x 0.0250 mol Cl- )/(50.0 ml x 1 L/103 mL) = 2.20 M

[Ca2+] = 0.0250 mol CaCl2/(50.0 ml x 1 L/103 mL) x 1 mol Ca2+/ 1 mol CaCl2 = 0.500 M