Question

A 25.0-mL sample of a 0.070 M solution of aqeous trimethylamine is titrated with a 0.088 M solution of HCl. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pKb of (CH3)3N = 4.19 at 25 C.

Answer #1

A 25.0 mL sample of a 0.0500 M solution of aqueous
trimethylamine is titrated with a 0.0625 M solution of HCl.
Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of
acid have been added; pKb of (CH3)3N = 4.19 at 25°C.

A 25.0 mL sample of a 0.290 M solution of aqueous trimethylamine
is titrated with a 0.363 M solution of HCl. Calculate the pH of the
solution after 10.0, 20.0, and 30.0 mL of acid have been added; pKb
of (CH3)3N=4.19 at 25 degrees C

A 25.0 ml sample of 0.090 M solution of aqueous trimethylamine
is nitrated with a 0.11 M solution of HCl. Calculate the pH of the
solution after 10.0, 20.0, and 30.0 mL of acid have been added, pkb
of (ch3)3n = 4.19 (part 2, after adding 20.0 mL)

A 25.0-mL sample of a 0.350 M solution of aqueous trimelhylamine
is titrated with a 0.438 M solution of HCI. Calculate the pH of the
solution after 10.0, 20.0, and 30.0 mL of acid have been added;
pK_b of (CH_3)_3N = 4.19 at 25 degree C. pH after 10.0 mL of acid
have been added: pH after 20.0 mL of acid have been added: pH after
30.0 mL of acid have been added:

A 25.0 mL sample of a 0.115 M solution of acetic acid is
titrated with a 0.144 M solution of NaOH. Calculate the pH of the
titration mixture after 10.0, 20.0, and 30.0 mL of base have been
added. (The Ka for acetic acid is 1.76 x 10^-5).
10.0 mL of base =
20.0 mL of base =
30.0 mL of base =

An
analytical chemist is titrating 81.1 mL of a 0.5900 M solution of
trimethylamine ((CH3)3N) with a 0.1300 M solution of HNO3. The pKb
of trimethylamine is 4.19. Calculate the pH of the base solution
after the chemist has added 415.2 mL of HNO3 solution to it. Round
your answer to 2 decimal places.

A 20.4 mL sample of 0.325 M
trimethylamine, (CH3)3N, is titrated
with 0.353 M perchloric acid. After adding
7.40 mL of perchloric acid, what will be
the pH ?

A 21.8 mL sample of 0.223 M trimethylamine, (CH3)3N, is titrated
with 0.207 M hydrobromic acid. At the equivalence point, the pH
is

A volume of 25.0 mL OF 0.140 M HCl is titrated against a 0.140 M
CH3NH2 solution added to it from a buret.
A.) Calculate the pH value of the solution after 10.0 mL of
CH3NH2 solution has been added.
B.) Calculate the pH value of the solution after 25.0 mL of
CH3NH2 solution have been added.
C.) Calculate the pH value of the solution after 35.0 mL of
CH3NH2 solution have been added.

a) 125 mL of a 0.098 M aniline (pKb = 9.40) solution is titrated
with 0.217 M HCl. What is the pH after 75 ml of HCl have been
added?
b) 125 mL of a 0.098 M aniline (pKb = 9.40) solution is titrated
with 0.217 M HCl. What is the pH after 25 ml of HCl have been
added?
c) 125 mL of a 0.098 M aniline (pKb = 9.40) solution is titrated
with 0.217 M HCl. What is...

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