When 17.28 mL of a 0.078 M aqueous solution of Na2SO4 is combined with 11.30 mL of a 0.20 M aqueous solution of NaCl and 7.84 mL of a 0.26 M aqueous solution of KCl, what is the total concentration of Na + in the combined solution?

[ Na + ]sln = 0.136 M

Explanation:

Na2SO4 → 2Na + (aq) + SO42 - (aq) NaCl → Na + (aq) + Cl - (aq) KCl → K + (aq) + Cl - (aq)

∴ mol Na2SO4 = (0.01728 L) * (0.078 mol/L) = 1.348 E-3 mol Na2SO4

⇒ mol Na + = (1.348 E-3 mol Na2SO4) * (2 mol Na+/mol Na2SO4)

⇒ mol Na + = 2.696 E-3 mol

∴ mol NaCl = (0.01130 L) * (0.20 mol/L) = 2.26 E-3 mol NaCl

⇒ mol Na + = (2.26 E-3 mol NaCl) * (mol Na+/mol NaCl) = 2.26 E-3 mol Na+

⇒ total moles Na + = 2.696 E-3 mol + 2.26 E-3 mol = 4.956 E-3 mol Na+

∴ total V sln = 17.28 mL + 11.30 mL + 7.84 mL = 36.42 mL = 0.03642 L sln

⇒ [ Na + ]sln = (4.956 E-3 mol) / (0.03642 L) = 0.136 M