Question

04 Question (a points) aSee page 166 Watch the ChemTour animation below on acid-base titrations. Then, answer the questions about the following reaction: H)2(aq) + 2HC1(aq)-, Caci2(aq) +H2O(1) An aqueous solution of Ca(OH)2with a concentration of 0.143 Mwas used to titrate 25.00 mL of aqueous HCI. 14.73 mL of the Ca(OH)2was required to reach the endpoint of the titration.

An aqueous solution of Ca(OH)₂with a concentration of 0.143 M was used to titrate 25.00 mL of aqueous HCl. 14.73 mL of the Ca(OH)₂was required to reach the endpoint of the titration.
How many moles of base were required to react completely with the acid in this reaction?

How many moles of HCl were present in the original 25.00 mL of acid?

What is the molarity of the original HCl solution?

Answer 1

Given,

Concentration of Ca(OH)₂ = 0.143 M

The volume of Ca(OH)₂ required to reach the endpoint = 14.73 mL = 0.01473 L

Also, Volume of HCl solution = 25.00 mL = 0.025 L

Now, The balanced chemical reaction between HCl and Ca(OH)₂ is,

2HCl_(aq) + Ca(OH)_2(aq)$\rightarrow$ CaCl_2(s) + 2H₂O_(l)

Calculating the number of moles of Ca(OH)₂ = 0.143 M x 0.01473 L =0.002106 moles of Ca(OH)₂

Thus, 0.00210 Or 0.00211 moles of base were required to react completely with the acid in this reaction.

Now, using the balanced chemical reaction and the moles of Ca(OH)₂, Calculating the number of moles of HCl required to react completely with Ca(OH)_2.

= 0.002106 moles of Ca(OH)₂ x ( 2 moles of HCl / 1 mole of Ca(OH)₂)

= 0.00421 moles of HCl

Thus, 0.00421 moles of HCl were present in the original 25.00 mL of acid.

Now, Calculating the molarity of HCl solution,

= 0.004213 moles of HCl / 0.025 L

= 0.169 M

The molarity of the original HCl solution is 0.169 M.