Question

6.4 mL of 0.625 M hydroiodic acid is added to 31.0 mL of barium hydroxide, and the resulting solution is found to be acidic. 27.9 mL of 0.760 M potassium hydroxide is required to reach neutrality. What is the molarity of the original barium hydroxide solution?

Answer 1

Molarity of Ba(OH)₂ is 0.227 M

Step-by-step explanation:

We are given;

• Volume of HCL as 56.4 mL (Assumption since HCl is in excess)
• Molarity of HCl as 0.625 M
• Volume of Ba(OH)₂ as 31.0 mL

At this point, we can determine the original number of moles of HCl and also write the equation between HCl and Ba(OH)₂.

Moles = Volume × Molarity

Therefore;

Moles of HCl given = 0.625 M × 0.0564 L

= 0.03525 moles

The equation for the reaction is;

Equation 1:

2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

• Further; we are given;

• Volume of KOH the reacted with excess HCl as 27.9 mL
• Molarity of KOH that reacted with excess HCL as 0.760 M

Therefore;

We can determine the moles of KOH that reacted with excess HCl and also write the equation for the reaction.

Moles = Molarity × volume

Molarity of KOH = 0.760 M × 0.0279 L

= 0.0212 moles

The equation for the reaction between HCl and KOH

Equation 2:

HCl + KOH → KCl + H₂O

• From the reaction we can determine the numb er of moles of HCl that reacted with KOH.

• Moles ratio of HCl to KOH is 1 : 1

Thus;

Moles of HCl = moles of KOH

Therefore, moles of HCl that reacted with KOH is 0.0212 moles

This means moles of HCl that reacted with Ba(OH)₂ will be given by;

= Original moles of HCl minus moles of HCl that reacted with KOH

= 0.03525 moles - 0.02120 moles

= 0.01405 moles

Thus, Moles of HCl that reacted with Ba(OH)₂ is 0.01405 moles

But,from the first equation;

Mole ratio of HCl to Ba(OH)₂ is 2 : 1

Thus;

Moles of Ba(OH)₂ = Moles of HCl ÷ 2

= 0.01405 moles ÷ 2

= 0.007025 moles

But, Molarity = Moles ÷ Volume

Molarity of Ba(OH)₂ = 0.007025 moles ÷ 0.031 L

= 0.227 M

Thus, the molarity of Ba(OH)₂ is 0.227 M