Question

a.)100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to the reaction flask. Classify the following conditions based on whether they are before the equivalence point, at the equivalence point, or after the equivalence point/endpoint.

1.) 150 mL of 1 M NaOH
2.) 200 mL of 1 M NaOH
3.) 50 mL of 1 M NaOH
4.) 100 mL of 1 M NaOH
5.) 5.00 mL of 1 M NaOH
6.) 10.0 mL of 1 M NaOH


b.)100.0 mL of a 0.510 M HNO_3 solution is titrated with 0.730 M KOH. Calculate the volume of KOH required to reach the equivalence point.

Answer 1

Concepts and reason

Equivalence point of a chemical reaction or some time also called end point, of chemical reaction. It’s mean it is a point at which chemically equivalent quantities of acids and bases have been mixed. In other way we can describe, the mole of acid mixed with the equivalently with the mole of bases. Where saturations points occurs means total moles of acid react with total moles of base no excess amount of any acid or base finally presents in that solutions.

Fundamentals

For the equivalent point, number of moles of a component will be equivalent; means total moles of acid need to react with total moles of base. It’s may be in different molar ratio. (This does not means or necessarily imply a 1 : 1 molar ratio of acid : base, merely that the ratio is the same as in balance chemical reactions equations). And it calculated by the equivalent moles of acid required for base so, it can be found by balance chemical reaction.

The titration contains HCl and NaOH.

Balance Chemical reaction.

From balance chemical titration reaction of HCl and NaOH.

The ratio of moles, of HCl and NaOH is 1:1 so, for the equivalent point,

1 mole of HCl required 1 Mole of NaOH,

Now, Find moles of HCl,

Part a.1

Find the mole of NaOH :

150 ml of 1.0 M NaOH

And the Mole of HCl is 0.1 mol and the mole of NaOH are 0.15 mol Therefore, the equivalence point will be “After equivalence point”

Part a.2

200 ml of 1.0 M NaOH

And the Mole of HCl is 0.1 mol and the mole of NaOH are 0.20 mol Therefore, the equivalence point will be “ After equivalence point”

[Part a.2]

Part a.3

50 ml of 1.0 M NaOH

And the Mole of HCl is 0.1 mol and the mole of NaOH are 0.05 mol Therefore, the equivalence point will be “Before equivalence point”

Part a.4

100 ml of 1.0 M NaOH

And the Mole of HCl is 0.1 mol and the mole of NaOH are 0.1 mol Therefore, the equivalence point will be “ At equivalence point”

[Part a.4]

Part a.5

5.00 ml of 1.0 M NaOH

And the Mole of HCl is 0.1 mol and the mole of NaOH are 0.005 mol Therefore, the equivalence point will be “ before equivalence point”

Part a.6

10.00 ml of 1.0 M NaOH

And the Mole of HCl is 0.1 mol and the mole of NaOH are 0.01 mol Therefore, the equivalence point will be “before equivalence point”

[Part a.6]

Part a.6

Part b

The titration contains KOH and

Balance Chemical reaction.

From balance chemical titration reaction of KOH and

The ratio of moles, of KOH and is 1:1 so, for the equivalent point,

1 mole of KOH required 1 Mole of

Now, Find moles of

Find the Volume of KOH :

0.051 mole and 0.730 M KOH

Volume of KOH required for the equivalent point with Is 69.86 ml ~ 70 ml

Ans: Part a.1

Therefore, the equivalence point “After equivalence point”

Part a.2

Therefore, the equivalence point “After equivalence point”

Part a.3

Therefore, the equivalence point “Before equivalence point”

Part a.4

Therefore, the equivalence point “At equivalence point”

Part a.5

Therefore, the equivalence point “Before equivalence point”

Part a.6

Therefore, the equivalence point “Before equivalence point”

Part b

Therefore, the volume of KOH “At equivalence point” = 70 ml