Question

Indicate the concentration of each ion present in the solution formed by mixing the following.

a) 15.0 mL of 0.300 M Na2SO4 and 20.4 mL of 0.100 M NaCl

Na+ :

SO42- :

Cl- :

b)3.50 g of KCl in 60.0 mL of 0.396 M CaCl2 solution (assume that the total volume is 60.0 mL)

K+ :

Ca2+ :

Cl- :

Answer 1

moles of Na2SO4 = Molarity x volume in liters = 0.3M x 0.015L = 0.0045 mol

each mole is having 2 mol of Na+ and one mole of SO42- accordingly

0.0045 mole Na2SO4 will have 0.0045 x 2 = 0.009 mol of Na+

0.0045 mole of Na2SO4 will have 0.0045 x 1 = 0.0045 mol of SO42-

moles of NaCl = 0.1 M x 0.0204 L = 0.00204 mol

0.00204 mol NaCl will have 0.00204 mol Na+ and 0.00204 mol Cl-

total volume = 15 + 20.4 = 35.4 mL = 0.0354 L

total moles of Na + = 0.009 + 0.00204 = 0.01104 mol

[Na+] = total moles of Na+ / total volume = 0.01104 mol / 0.0354 L = 0.312 mol / L

[SO42-] = 0.0045 mol / 0.0354 L = 0.127 mol /L

[Cl-] = 0.00204 mol / 0.0354 L = 0.0576 mol /L