Question

If 100. mL of 0.200 M Na2SO4 is added to 200. mL of 0.300 M NaCl, what is the concentration of Na+ ions in the final solution? Assume that the volumes are additive.

Answer 1

number of moles of Na+ from Na2SO4 = 2*M*V
                          = 2*0.2 M * 0.1 L
                          = 0.04 mol

number of moles of Na+ from NaCl = M*V
                          = 0.3 M * 0.2 L
                          = 0.06 mol

Total number of moles of Na+ = 0.04 + 0.06 = 0.1 mol

total volume = 100 mL + 200 mL = 300 mL = 0.3 L

[Na+] = number of moles / volume
             = 0.1 / 0.3
             =0.33 M
Answer: 0.33 M