Question

What is the pH of a 0.084 M solution of hydrazoic acid. HN3?

Answer #1

its a week acid we need to construct the ICE table

since water is exceaa we do not to consider the concentration of H2O

HN_{3} + H_{2}O --->
H_{3}O^{+} + N3^{-}

I 0.084 0 0

C -x +x +x

E 0.084-x +x +x

now write the equilibrium expression or dissociation constant

Ka = [N3^{-} ] [H_{3}O^{+}] /
[HN_{3} ]

Ka value for the hydrozoic acid is 2.5 x 10^{-5} (from
the text book standard value)

2.5 x 10^{-5} = [x][x] / [0.084-x]

x^{2} + 2.5 x 10^{-5}x -
2.1x10^{-6} = 0

solve the quadratic equation

x = 0.001436 = [H3O+]

pH = -log[H3O+]

pH = -log[0.001436]

pH = 2.84

The Ka of hydrazoic acid (HN3) is 1.9 x 10-5 at 25.0 °C. What is
the pH of a 0.40 M aqueous solution of HN3?

A solution is labeled "0.462 M HN3." What are the following in
the solution? Express your answers to three significant figures
even when only two are valid. Acid-base table [HN3] = M [H3O1+] = M
[N31-] = M pH =

Calculate the percent ionization of hydrazoic acid (HN3)
in solutions of each of the following concentrations (Ka
is given in Appendix D in the textbook). (I dont own the textbook
but I found the Ka of HN3 to be 1.9*10^-5)
Part A
0.414 M .
Express your answer using two significant figures.
________ %
Part B
0.118 M .
Express your answer using two significant figures.
________ %
Part C
4.16×10−2 M .
Express your answer using two significant figures....

Calculate the percent ionization of hydrazoic acid (HN3) in
solutions of each of the following concentrations (Ka is given in
Appendix D in the textbook). Ka is
1.9x10-5
A). 0.383M
B). 0.113M
C). 3.66x10-2

Problem 16.63
Calculate the percent ionization of hydrazoic acid (HN3) in
solutions of each of the following concentrations (Ka is
given in Appendix D in the textbook).
Part A
0.388 M .
Express your answer using two significant figures.
%
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Up
Part B
0.117 M .
Express your answer using two significant figures.
%
SubmitMy AnswersGive
Up
Part C
3.61×10−2 M .
Express your answer using two significant figures.
%

A solution is prepared by dissolving 0.23 mol of hydrazoic acid
and 0.27 mol of sodium azide in water sufficient to yield 1.00 L of
solution.The addition of 0.05 mol of NaOH to this buffer solution
causes the pH to increase slightly. The pH does not increase
drastically because the NaOH reacts with the ________ present in
the buffer solution. The Ka of hydrazoic acid is 1.9 × 10-5.

Calculate the pH at the equivalence point in titrating 0.090 M
solutions of each of the following with 0.064 M NaOH.
(a) hydroiodic acid (HI)
pH =
(b) hydrazoic acid (HN3), Ka = 1.9e-05
pH =
(c) arsenous acid (H3AsO3), Ka =
5.1e-10
pH =

Consider the titration of a 45.0 mL sample of 0.100 M HN3 with
0.250 M KOH. (Ka of HN3 = 2.5 x 10-5) Calculate each of the
following: (a) How many mL are required to reach the equivalence
point?
(b) What is the initial pH of the acid solution?
(c) What is the pH after the addition of 5.00 mL of KOH?
(d) What is the pH after the addition of 9.00 mL of KOH? (e)
What is the pH...

What is the pH of a 0.35 M solution of 3-Chloropropanoic Acid?
pH=
What is the pH of a 0.47 M solution of the base Benzylamine?
pH=

What is the pH of a 0.13 M solution of hypoiodous acid?

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