Question

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

(a) 30 mL of 0.100 M HCl and 10.0 mL of 0.520 M HCl

H+ _____M

Cl-______ M

(b) 15.0 mL of 0.294 M Na2SO4 and 28.2 mL of 0.200 M KCl

Na+______ M

K+______ M

SO42-______ M

Cl-______ M

(c) 3.50 g of NaCl in 59.3 mL of 0.439 M CaCl2 solution

Na+______ M

Ca2+______ M

Cl-______ M

Answer 1

a.)

Mole HCL = 30 x 0.100 / 1000 = 0.003
Mole HCl = 10 x 0.520 / 1000 = 0.0052
Total mole = 0.003 + 0.0052 = 0.0082
Total volume = 30 + 10 = 50 mL = 0.040 L
HCl is a strong acid >> H+ + Cl-
[H+] = [Cl-] = 0.0082 / 0.040 = 0.205 M

b.)

Mole Na2SO4 = 15 x 0.294 / 1000 = 0.00441
Na2SO4 >> 2Na+ + SO42-
Mole Na+ = 2 x 0.00441 = 0.00882
Mole SO42- = 0.00441
Mole KCl = 28.2 x 0.200 /1000 = 0.00564
KCl >> K+ + Cl-
Mole K+ = mole Cl- = 0.00564
Total volume = 15 + 28.2 = 43.2 mL = 0.0432 L
[Na+] = 0.00882 /0.0432 = 0.204 M
[SO42-] = 0.00441 / 0.0432 = 0.102 M
[K+] = [Cl-] = 0.00564 / 0.0432 = 0.131 M

c.)

c)
Number of moles of NaCl = 3.5/(35.5+23)=3.5/58.5 = 5.98 *10^-2
Number of moles of Na+ = 5.98 *10^-2 moles = 0.0598 moles
Total number of moles of Cl- = 2*0.439 * 0.0593 + 5.98*10^-2 = 2*0.0260327+0.0598 = 0.112 moles

Molarity of Na+ = 0.0598*1000/59.3 = 1.008 M
Molarity of Ca+2 = 0.439 M
Molarity of Cl- = 0.112*1000/59.3 = 1.888 M