How many moles of acetic acid remain unreacted whenever 20 ml of .1 M Ba(OH)2 is added to 50 ml .1 M acid?

Question

How many moles of acetic acid remain unreacted whenever 20 ml of .1 M Ba(OH)2 is added to 50 ml .1 M acid?

Answer 1

First you need a balanced equation so you can see what mole ratio the reactants combine in.

2(CH3COOH) + Ba(OH)2 ------------------> (CH3COO)2Ba + 2H2O

Now work out the number of moles of each reactant are in the reaction mixture. Use the equation:

Molarity = moles / litres

moles of Ba(OH)2 in 20 ml 0.1M Ba(OH)2

Molarity = moles / litres

0.10 M = moles / 0.020 L

therefore moles = 0.10 M x 0.020 L

=0.002 moles of Ba(OH)2

moles 50 ml of 0.1 M acetic acid

Molarity = moles / Litres

0.1M = moles / 0.050 L

therefore moles = 0.1 M x 0.050 L

= 0.005 moles of acetic acid

From the balanced equation, 2 moles of acetic acid is required to react with 1 mol of Ba(OH)2. You have 0.002 moles of Ba(OH)2 therefore you need

2 x 0.002 moles of Acetic acid. = 0.004 moles

You have 0.005 moles of acid in 50 ml of 0.1 M acid

therefore 0.001 mole is left unreacted.