Question

if 400 mL of 0.20 M HCl solution is added to 800 mL of 0.0500 M NaOH solution, the resulting solution will be _______ in sodium ions

Answer 1

no of moles of HCl   = molarity * volume in L

                             = 0.2*0.4   = 0.08moles

no of moles of NaOH = molarity * volume in L

                                = 0.05*0.8 = 0.04moles

HCl(aq) + NaOH(aq) --------------> NaCl(aq) + H2O(l)

1 mole of HCl react with 1 mole of NaOH

0.08 moles of HCl react with 0.08 moles of NaOH is required

NaOH is limiting reactant

'1 mole of NaOH react with excess of HCl to gives 1 mole of NaCl

0.04moles of NaOH react with excess of HCl to gives 0.04 moles of NaCl

total volume of solution = 800 +400 = 1200ml = 1.2L

molarity of NaCl = no of moles/total volume of solution in L

                          = 0.04/1.2   = 0.033M

NaCl(aq) -----------> Na^+ (aq) + Cl^-(aq)

0.034M ---------------0.034M

0.034M of Na^+ ions >>>>>answer