Question

75.0 mL of 0.250 M hydrochloric acid is added to 225.0 mL of 0.0550 M Ba(OH)₂ solution. What is the concentration of Ba²⁺ ion left in solution?

75.0 mL of 0.250 M hydrochloric acid is added to 225.0 mL of 0.0550 M Ba(OH)₂ solution. What is the concentration of Cl^- ion left in solution?

Answer 1

Ba(OH)2 (aq) + 2HCl(aq) ------------------> BaCl2(aq) + 2H2O(l)

no of moles of Ba(OH)2 = molarity * volume in L

                                        = 0.055*0.225   = 0.012375 moles

no of moles of HCl         = molarity * volume in L

                                        = 0.25*0.075   = 0.01875 moles

2 moles of HCl react with 1 mole of Ba(OH)2

0.01875 moles of HCl react with = 1*0.01875/2    = 0.009375 moles of Ba(OH)2

Ba(OH)2 is excess reactant

no of moles of excess of Ba(OH)2 = 0.012375-0.009375   =0.003 moles of Ba(OH)2

total volume = 75+225 = 300ml = 0.3L

molarity of Ba(OH)2   = no of moles/total volume in L

                                 = 0.003/0.3 = 0.01 M of Ba(OH)2 is left form the solution

0.01M Ba^2+ is left from the solution.

1 moles of Ba(OH)2 react with 2 moles of HCl

0.012375 moles of Ba(OH)2 react with = 2*0.012375/1 = 0.02475 moles of HCl is required

    HCl is limiting reactant

All HCl is completed participate in the reaction

The conc of Cl^- left is zero.