Thirteen Two 25.0mL sample, one 0.100 M HCI and the other 0.100 M HF, were titra

Question

Thirteen

Two 25.0mL sample, one 0.100 M HCI and the other 0.100 M HF, were titrated with .200 MKOH. Answer each of the following questions regarding these two titrations.

Part A

What is the volume of added base at rhe equilalence point for HCl?

Part B

What is the volume of added base at the equilance piont for HF?

Part CPredict whether the pH at the equivalence piont for each will be acidic, basic, or neutral.?

Part D

Predict which titration curve will have the lower intial pH.

neutral for HCl, and acidic for HF

neutral for HCl, and basic fro HF

neutral for HF, and acidic for HCl

neutral for HF, and basic for HCl

Explanation / Answer

A)25 mL of 0.100M HVCl is titrated with 0.200M KOH

At equivalence mmoles of HCl = mmoles of KOH

25mL x 0.100M = VmL x 0.200M

Thus volume of KOH at equivalence = 12.5 mL

Part B )

25 mL of 0.100M HF is titrated with 0.200M KOH

At equivalence mmoles of HF = mmoles of KOH

25mL x 0.100M = VmL x 0.200M

Thus volume of KOH at equivalence = 12.5 mL

Part C)

HCl is a strong acid and KOH is a strong base.

On neutralisation

HCl + KOH -----------> KCl + H2O

the salt formed is a salt of strong acid and strong base KCl, which does not hydrolyze in water.

Thus the solution is neutral at equivalence.

HF is a weak acid and KOH is a strong base.

On neutralisation

HF + KOH -----------> KF + H2O

the salt formed is a salt of weak acid and strong base KF, which hydrolyzes in water to give a basic solution.

Thus the solution is basic at equivalence.

Part D

As Hcl is a strong acid and HF is a weak acid, at same concentration of acids, the [H3O+] produced in HF is less than that of [3O+] in HCl. Thus th einital pH of HCl is lower than that of HF .

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