Question

56.7 mL of 1.18 M hydrochloric acid is added to 28.7 mL of barium hydroxide, and the resulting solution is found to be acidic.
21.3 mL of 1.27 M sodium hydroxide is required to reach neutrality.
what is the molarity of the original calcium hydroxide solution? __M

Answer 1

Concepts used to solve:

· Molarity =

· To find how much substance is produced or consumed, always compare the number of moles from the balanced chemical reaction.

So, no. of moles of NaOH added = 1.27 M x 0.0213 L = 0.027 moles

Total no. of moles of HCl in solution = 1.18 M x 0.0567 L = 0.067 moles

When HCl is added to barium hydroxide (Ba(OH)2) the following reaction takes place:

2HCl + Ba(OH)2 = BaCl2 + 2H2O

HCl is acidic, Ba(OH)2 is a base. BaCl2 is a salt. So if the resulting solution is acidic, then it means that HCl is left over unreacted when all the Ba(OH)2 has been used up.

NaOH is then added to react with HCl and make the solution neutral. The reaction is:

NaOH + HCl = NaCl + H2O

So, to get the molarity of the original Ba(OH)2 solution, we first need to find out the number of moles of Ba(OH)2 in the solution by the following steps:

· From the reaction between NaOH and HCl we can find the excess no. of moles of HCl that was left over.

· Since we know the molarity and volume of HCl added, we know the total no. of moles of HCl in the solution.

· If we subtract the excess no. of moles from the total moles of HCl, we get the no. of moles of HCl which reacted with Ba(OH)2 , and thus, find the no. of moles of BaCl2 present initially.

To find excess moles of HCl:

No. of moles of NaOH added = 0.027 moles

From the equation, 1 mole NaOH reacts with 1 mole HCl.

So, 0.027 moles NaOH reacted with 0.027 moles HCl.

Thus, excess moles of HCl = 0.027 moles.

Total moles of HCl in solution = 0.067 moles

So, no. of moles of HCl which had reacted with Ba(OH)2 = 0.067 – 0.027 = 0.04 moles

From the reaction between HCl and Ba(OH)2,

2 mole HCl reacts with 1 mole Ba(OH)2.

So, 0.04 moles HCl reacts with ½ x 0.04 moles = 0.02 moles Ba(OH)2.

Thus, no. of Ba(OH)2 added to solution = 0.02 moles

According to the question, 0.02 moles were present in 28.7 mL of Ba(OH)2 .

So, molarity of the solution = 0.02 moles/0.0287 L = 0.7 M