Question

a.) What is the chloride ion concentration in the solution prepared by combining 10.00 mL of 0.150 M NaCl solution, 15.00mL of 0.125 M CaCl2 solution and 25.00 mL of 0.110 M AlCl3 solution?

b.) What is the chloride ion concentration in the solution made by adding 125.0 mL of water to 17.00 mL of the solution prepared in part A?

*Please explain as best as possible so I can get a better understanding of what it is asking for, Thank you!

Answer 1

1)Molar mass BaCl2 = 208.2324 g/mol

Moles BaCl2 = 45.0 / 208.2324 = 0.216
1- M = 0.216 / 0.750 = 0.288 M
2- 2 x 0.288 = 0.576 M
3 - Mass water = 748 g
Molar mass water = 18.02 g/mol
Moles water = 748 / 18.02 = 41.5
Molecules of water = 41.5 x 6.02 x 10^23 = 2.50 x 10^25
Moles Cl- = 0.576 x 750 / 1000 = 0.432
Number of Cl- = 0.432 x 6.02 x 10^23 = 2.60 x 10^23
Molecules of H2O for 1 ion Cl- =2.50 x 10^25 / 2.60 x 10^23 =
= 96
4- 0.0034 x 0.500 = 0.288 V
V = 0.00590 = 5.90 mL


Moles NaOH = 10.0 / 40 = 0.25
Moles H2SO4 = 0.25 / 2 = 0.125
V = 0.125 / 0.100 = 1.25 L = 1250 mL

2)  First calculate the moles of NaCl using g NaCl.
Molar mass of NaCl = 58.45 g
Mole NaCl = 1.60 g NaCl ( 1 mole NaCl/ 58.45 g NaCl) = 0.0274 mole NaCl (answer to 3 sig.figs)

Next calculate the moles of CaCl2 using M of CaCl2 and V (in L) of CaCl2.
moles CaCl2 = (M) (L)
= (0.100 M) (0.0500L)
= (0.100 moles/L) (0.0500 L) = 0.00500 moles CaCl2
Find the concentration of Cl- ion from NaCl and CaCl2 separately and add the two to get mole Cl-.

NaCl(aq) --> Na+(aq) + Cl-(aq)
0.0274 mole NaCl (1mole Cl-/1mole naCl) = 0.0274 mole Cl-

CaCl2(aq) --> Ca^2+(aq) + 2 Cl-(aq)
0.00500 mole CaCl2 ( 2 mole Cl-/1 mole CaCl2) = 0.0100 mole Cl-

Total concentration of Cl- = 0.0274 mole Cl- + 0.0100 mole Cl-
= 0.0374 mole Cl-

Molarity of Cl- = moles Cl-/L of solution
= 0.0374 mole / 0.0500 L = 0.748 M