Select the single best answer. A 50.0-mL quantity of a 020 M solution of one of

Question

Select the single best answer. A 50.0-mL quantity of a 020 M solution of one of the following weak bases is titrated with 0.050 M HCl. At the equivalence point, the pH is 2.99. Identify the weak base. A. ethylamine (Kb = 5.6 x 10^-4) B. methylamine (Kb = 4.4 x 10^-4) C. ammonia (Kb = 1.8 x 10^-5) D. pyridine (Kb = 1.7 x 10^-9) E. aniline (Kb = 3.8 x 10^-10) F. urea (Kb = 1.5 x 10^-14) Be sure to answer all parts. Which indicators that would be suitable for each of the following titrations: (a) CH3NH2 with HBr (b) HNO3 with NaOH (c) HNO2 with KOH

Explanation / Answer

If this is a weak base, this will most likely form hydrolysis

the base will form a conjugate acid, which is acidic...

BH+ + H2O  <-> B + H3O+

Ka = [B][H3O+]/[BH+]

mmol of base = 50*0.2 = 10 mmol

V acid required = mmol/M = 10/0.05 = 200 mL

Total V = 200+10 = 210 mL

[B] = 10/210 = 0.04761 M

Ka = [B][H3O+]/[BH+]

n equilbirium

[B] = [H3O+] = x

pH = 2.99

[H+] = 10^-pH = 10^-2.99 = 0.001023 M

x = 0.001023

[BH+] = 0.04761 -0.001023 = 0.046587

Ka = (0.001023 )(0.001023 )/(0.046587)

Ka = 0.00002246

Ka = 2.2*10^-5

Kb = (Kw/Ka) = (10^-14)/(2.2*10^-5) = 4.54*10^-10

best answer isANILINE

b)

CH3NH2 has a pKb:

Identify the pH in equivalnece:

pH < 7 i.e pH 4-6. So the best indicator is methyl red

which shows color at pH 4 -6.

b)

HNO3 is strong acid; NaOH = strong base

pH = neutral

So use any range of:

pH > 7

phenol red, color change at pH 7-9

even phenolphthalein, since a very small amount of excess NaOh will increase pH drastically

c)

HNO2 --> weak acid

once again, expect a basic pH, since

NO2- + H2O <-> HNO2 + OH-

This will have a higher pH

use phenolphthalein

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