Answer to Question #73726 in General Chemistry for aleia

Question

Indicate the concentration of each ion present in the solution formed by mixing the following.
(a) 12 mL of 0.100 M HCl and 10.0 mL of 0.220 M HCl
H+ :
M
Cl- :
M

(b) 15.0 mL of 0.300 M Na2SO4 and 28.0 mL of 0.100 M NaCl
Na+ :
M
SO42- :
M
Cl- :
M

(c) 3.50 g of KCl in 60.0 mL of 0.316 M CaCl2 solution (assume that the volumes are additive)
K+ :
M
Ca2+ :
M
Cl- :
M

Answer 1

(a) 12 mL of 0.100 M HCl and 10.0 mL of 0.220 M HCl

n1(HCl) = C*V = 0.012L*0.1mol/L = 0.0012 mol

n2(HCl) = C*V = 0.01L*0.22mol/L = 0.0022 mol

HCl → H+ + Cl-

n1(H+) = 0.0012 mol

n2(H+) = 0.0022 mol

n1(Cl-) = 0.0012 mol

n1(Cl-) = 0.0022 mol

C(H+) = (0.0012 + 0.0022)mol / (0.012 + 0.01)L = 0.0034mol/0.022L = 0.1545 M

C(Cl-) = (0.0012 + 0.0022)mol / (0.012 + 0.01)L = 0.0034mol/0.022L = 0.1545 M

(b) 15.0 mL of 0.300 M Na2SO4 and 28.0 mL of 0.100 M NaCl

n(Na2SO4) = C*V = 0.015L*0.3mol/L = 0.0045 mol

n1(NaCl) = C*V = 0.028L*0.1mol/L = 0.0028 mol

Na2SO4 → 2Na+ + SO42-

NaCl → Na+ + Cl-

n1(Na+) = 0.0045*2 = 0.009 mol

n2(Na+) = 0.0028 mol

n(Cl-) = 0.0028 mol

n(SO42-) = 0.0045 mol

V = 0.015 + 0.028 = 0.043 L

C(Na+) = (0.009+0.0028)mol/0.043L = 0.2744 M

C(Cl-) = 0.0028 mol/0.043L = 0.0651 M

C(SO42-) = 0.0045 mol/0.043L = 0.1047 M

(c) 3.50 g of KCl in 60.0 mL of 0.316 M CaCl2 solution

n(CaCl2) = C*V = 0.06L*0.316mol/L = 0.019 mol

n(KCl) = 3.5g/74.5g/mol = 0.047 mol

KCl → K+ + Cl-

CaCl2 → Ca2+ + 2Cl-

n(Ca2+) = 0.019 mol

n(Cl-) = 0.019*2 +0.047 = 0.085 mol

n(K+) = 0.047 mol

C(Ca2+) = 0.019 mol/0.06 L = 0.317 M

C(Cl-) = 0.085 mol/0.06 L = 1.417 M

C(K+) = 0.047 mol/0.06 L = 0.783 M