Question

Calculate the pH at the equivalence point for the following
titration:

0.1 M HCl and 0.1 M NH3

The finale answer should be 5.28

And there is no volume

Answer #1

0.1 M of HCl and NH3 will react to form 0.1 M of NH4+

find Ka of NH4+

use:

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/1.8*10^-5

Ka = 5.556*10^-10

NH4+ + H2O -----> NH3 + H+

0.1 0 0

0.1-x x x

Ka = [H+][NH3]/[NH4+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.556*10^-10)*0.1) = 7.454*10^-6

since c is much greater than x, our assumption is correct

so, x = 7.454*10^-6 M

so.[H+] = x = 7.454*10^-6 M

use:

pH = -log [H+]

= -log (7.454*10^-6)

= 5.13

Answer: 5.13

Calculate the pH at the equivalence point for the following
titration: 0.40 M HCl versus 0.40 M methylamine (CH3NH2). The Ka of
methylammonium is 2.3 × 10^−11

Calculate the pH at the equivalence point for the following
titration: 0.40 M HCl versus 0.40 M methylamine (CH3NH2). The Ka of
methylammonium is 2.3 × 10^−11

Calculate the pH at the equivalence point for the following
titration: 0.35 M HCl versus 0.35 M methylamine (CH3NH2). The Ka of
methylammonium is 2.3 × 10−11.

A. What is the pH at the equivalence point of a titration of
0.050 M HClO4 with 0.025 M Sr(OH)2? Why should this be an easy
question!?
B. A 25.0 mL sample of 1.44 M NH3 is titrated with 1.50 M HCl.
Choose an appropriate indicator for this titration. Justify your
answer.

Calculate the pH at the equivalence point for the titration of
0.180 M methylamine (CH3NH2) with 0.180 M HCl. The Kb of
methylamine is 5.0× 10–4.
pH=?

1. For the titration of 25.0 mL of 0.1 M HCl (aq) with 0.1 M
NaOH (aq), at what volume of NaOH (aq) should the equivalence point
be reached and why? If an additional 3.0 mL of 0.1 M NaOH (aq) is
then added, what is the expected pH of the final solution?
2. What is the initial pH expected for a 0.1 M solution of
acetic acid? For the titration of 25.0 mL of 0.1 M acetic acid with...

Calculate the pH at the equivalence point for the titration of
0.130 M methylamine (CH3NH2) with 0.130 M HCl. The Kb of
methylamine is 5.0× 10–4.

Calculate the pH at the equivalence point in the titration of
50.0 mL of 0.125 M methylamine (Kb = 4.4 × 10−4) with 0.265 M
HCl..

Calculate the pH at the equivalence point for the titration of
0.240 M methylamine (CH3NH2) with 0.240 M HCl. The Kb of
methylamine is 5.0× 10–4.

Calculate the pH at the equivalence point for the titration of
0.110 M methylamine (CH3NH2) with 0.110 M HCl. The Kb of
methylamine is 5.0× 10–4.

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 2 minutes ago

asked 21 minutes ago

asked 31 minutes ago

asked 42 minutes ago

asked 47 minutes ago

asked 52 minutes ago

asked 53 minutes ago

asked 58 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago