This is our balanced equation (note: all of the compounds except for water, like #HCl#, would usually be written as constituent ions (like #H^+# and #Cl^-#) since they dissociate in water, but we're just using formulas like #HCl (aq)# because they're easier to read):
#2HCl (aq) + Ba(OH)_2(aq) -> 2H_2O(l) + BaCl_2(aq)#
If all of the #OH^-# and #H^+# ions react, the concentration of #OH^-# ions will be #0#—there won't be anything left. But in our case, #Ba(OH)_2# is in excess, so not all #OH^-# ions will react and there still will be some left after the reaction is complete.
To find the concentration of #OH^-# ions, we need to find how many #OH^-# ions didn't react, and then use that to calculate its concentration.
Finding Excess #OH^-#
#2HCl (aq) + Ba(OH)_2(aq) -> 2H_2O(l) + BaCl_2(aq)#
For every #1# mole of #Ba(OH)_2# that reacts, #2# moles of #HCl# need to react.
We have #"20 mL"# of #"0.05 M HCl"#.
That's #"0.05 mol/L × 0.020 L = 0.001"# moles of #HCl#.
We have #"30 mL"# of #"0.10M Ba(OH)₂"#.
That's #"0.10 mol/L × 0.030 L = 0.003"# moles of #Ba(OH)_2#.
To react with #0.001# moles of #HCl#, #0.001xx2=0.002# moles of #Ba(OH)_2# need to react because of the mole ratio.
We know that we have #0.003# moles, so #0.003-0.002=0.001# moles of #Ba(OH)_2# didn't react.
This means that after the reaction is over, #0.001# moles of #Ba(OH)_2# are still left sitting there in the solution.
We can use this to find out how many #OH^-# are left in the solution: #1# mole of #Ba(OH)_2# contains #2# moles of #OH^-#, so #0.001# moles of #Ba(OH)_2# contains #0.002# moles of #OH^-#.
Calculating Concentration
Molarity is expressed in moles per litre:
#M = "mol"/L#
We know the moles of #OH^-#, which is #0.002#.
The volume in litres is also given to us—there are #20 mL# of #HCl# and #30 mL# of #Ba(OH)_2#, so there must be #20+30 = 50 mL#, or #0.05 L#, of total solution.
Therefore, the molarity is #"0.002 mol/0.05 L = 0.04M"#.
