A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of

Question

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the addition of 400.0 mL of KOH. The K_a of HF is 3.5 times 10^-4. 13.08 12.60 13.85 12.30 12.78 A 100.0 mL sample of 0.10 M Ca(OH)_2 is titrated with 0.10 M HBr. Determine the pH of the solution after the addition of 300.0 mL HBr. 1.60 1.30 1.00 12.40 1.12 When titrating a strong monoprotic acid and KOH at 25 degree C, the pH will be less than 7 at the equivalence point. pH will be greater than 7 at the equivalence point. titration will require more moles of base than acid to reach the equivalence point. pH will be equal to 7 at the equivalence point. titration will require more moles of acid than base to reach the equivalence point.

Explanation / Answer

25) the reaction is

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

100x0.2 400x0.1 0 0 initial milliequivalents

= 20 =40

0 20 20 20 equilibrium meq

0 20/500 20/500 20/500 equilibrium concentrations

Now the solution has a salt and a strong base. thus the pH of the solution is decided by the concentration of the base.

Now [OH-] = 20/500M

thus pH =14 -pOH= 14 - (- log [4x10-2])

= 12.6090

Option B is right.

26) Ca(OH)2 +2 HBr ---------> CaBr2 + 2H2O

100x2x0.1 300x0.1 0 0 initial milliequivalents

= 20 =30

0 10 20 20 equilibrium milliequivalents

0 10/400 20/400 20/400 equilibrium concentrations

Thus the [H+] = 10/400 N = 10/400M

Since the reamining solution has only the strong acid HBr nd salt CaBr2 , the pH of the solution is given by

pH = - log [H+] = - log 2.5 x10-2

= 1.6021

Thus option A is correct.

27) KOH is a strong monoacidic base , titrating with a strong monoprotic acid , at the equivlence point , the number of moles of acid reacted are equal to moles of base.

The pH of the solutio will be equal to 7, as the salt formedis a salt of strong acid and strong base wich does not undergo hydrolysis.

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