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:

∴ 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

the dependant variable is spilled water

the independent is the heat