Question

25.00 cm³ of the hydrochloric acid reacted with 23.50 cm³ of the 0.100 mol/dm³

barium hydroxide solution.
The equation for the reaction is:
2 HCl(aq) + Ba(OH)2(aq) → BaCl₂(aq) + 2H₂O(1)
Calculate the concentration of the hydrochloric acid in mol/dm³.

Please help!!

Answer 1

The concentration of the hydrochloric acid is 0.00047 mol/dm³.

Step-by-step explanation:

The balanced equation for the reaction is:

2 HCl(aq) + Ba(OH)2(aq) → BaCl₂(aq) + 2H₂O

We know that 25.00 cm³ of the hydrochloric acid reacted with 23.50 cm³ of the 0.100 mol/dm³ barium hydroxide solution.

Given the balanced equation, it is clear that 2 moles of hydrochloric acid react with 1 mole of barium hydroxide. As we know the volume and concentration of the barium hydroxide, we can calculate the number of moles of barium hydroxide present in the 23.50 cm³ of solution.

23.50 cm³ of 0.100 mol/dm³ barium hydroxide = (23.50 x 0.100) / 1000 = 0.00235 moles

As 2 moles of hydrochloric acid react with 1 mole of barium hydroxide, and we know the number of moles of barium hydroxide, we can calculate the number of moles of hydrochloric acid.

0.00235 moles of barium hydroxide / 2 = 0.001175 moles hydrochloric acid

We have the number of moles of hydrochloric acid, we can calculate the concentration in mol/dm³.

0.001175 moles hydrochloric acid / 25.00 cm³ = 0.00047 mol/dm³

The concentration of the hydrochloric acid is 0.00047 mol/dm³.

Answer 2

0.111 moldm⁻³

First work out the moles of NaOH dispensed from the burette:

1. (25.0/1000)x0.100 = 0.00250 moles of HCl

Next use the balanced equation to reason how many moles of HCl are present in the flask:

2. 1:1 therefore 0.00250 moles NaOH

Lastly, now that you have both the volume and the number of moles of HCl, work out its concentration.

3. 0.0025/(22.5/1000) = 0.111 moldm⁻³

Step-by-step explanation: