The pH of a 7.87×10 -1 M solution of a weak monoprotic acid is 3.96. Calculate t

Question

The pH of a 7.87×10-1 M solution of a weak monoprotic acid is 3.96. Calculate the percent dissociation of the acid to 3 significant figures.

The pH of a 7.72×10-2 M solution of a weak base that will accept only one proton is 7.93. Calculate the percent dissociation of the base to 3 significant figures.

Can the five percent approximation be applied in determining the pH of the following solutions?
A table of pKa values can be found here.
Yes No  7.36×10-5 M hypochlorous acid
Yes No  3.55×10-5 M dichloroacetic acid
Yes No  3.57×10-3 M trifluoroacetic acid
Yes No  9.87×10-3 M lactic acid

Predict whether aqueous solutions of the salts will be acidic, basic or neutral.
Rb(ClO4) - KHSO3 - C6H5NH3Br - Ca(ClO4)2 - NaC3H7COO

A sample of tris(hyroxymethyl)aminomethane (H2NC(CH2OH)3), which is a base, with a mass of 0.2716 g is dissolved in water and used to standardize a solution of perchloric acid. An endpoint is reached when 28.05 mL of perchloric acid has been added. Determine the molarity (in mol/L) of the perchloric acid solution. Report your answer to four significant figures.

The standardized perchloric acid solution is then used to titrate 22.11 mL of a solution of sodium lactate (NaC2H5OCOO). An endpoint is reached when 23.94 mL of perchloric acid has been added. Determine the molarity (in mol/L) of the sodium lactate solution. Report your answer to four significant figures.

Explanation / Answer

A) pH = 3.96 = -log[H+]

[H+] = 1.096 x 10^-4 M

percent dissociation = (1.096 x 10^-4/7.87 x 10^-1) x 100 = 0.014%

B) pH = 7.93

pOH = 14 - pH = 6.07

[OH-] = 8.51 x 10^-7 M

percent dissociation = (8.51 x 10^-7/7.72 x 10^-2) x 100 = 0.001%

C) five percent approximation

7.36 x 10^-5 M hypochlorous acid : Yes

It is a weak acid and so the rule applied to it.

3.55 x 10^-5 M dichloracetic acid : No

It is a strong acid and the rule fails to apply here

3.57 x 10^-3 M trifloroacetic acid : No

Very strong acid, the rule fails to apply here

9.87 x 10^-3 M lactic acid : Yes

It is again a weak acid so the rule works with it.

D) Aqueous solutions of,

salts of conjugate acid and conjugate base of strong base and strong acid are neutral

salt of conjugate acid of a strong base and conjugate base of a weak acid are basic

salt of conjugate acid of weak base and conjugate base of strong acid are acidic

salts of weak acid and weak base varies with their pKa values

So, we would have,

Rb(ClO4) : Neutral solution

KHSO3 : Neutral solution

C6H5NH3Br : acidic solution

Ca(ClO4)2 : Neutral solution

NaC3H7COO : Basic solution

E) 1 mole of base reacts qith 1 mole of HClO4

moles of base = 0.2716/121.14 = 2.24 x 10^-3 mols = moles of HClO4

molarity of acid = 2.24 x 10^-3/0.02805 = 0.07986 M

Now, moles of HClO4 used for sodium lactate = 0.08 x 0.02394 = 1.91 x 10^-3 mols

moles of sodium lactate = 1.91 x 10^-3/0.02211 = 0.086467 M

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