Question

5. Calculate pH of 0.25M aniline
(C_{6}H_{5}NH_{2}) solution.

Answer #1

C6H5NH2 dissociates as:

C6H5NH2 +H2O -----> C6H5NH3+ + OH-

0.25 0 0

0.25-x x x

Kb = [C6H5NH3+][OH-]/[C6H5NH2]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((4.3*10^-10)*0.25) = 1.037*10^-5

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

so, x = 1.037*10^-5 M

So, [OH-] = x = 1.037*10^-5 M

use:

pOH = -log [OH-]

= -log (1.037*10^-5)

= 4.9843

use:

PH = 14 - pOH

= 14 - 4.9843

= 9.0157

Answer: 9.02

Part A Find the [OH−] of a 0.45 M aniline (C6H5NH2) solution.
(The value of Kb for aniline (C6H5NH2) is 3.9×10−10.) Express your
answer to two significant figures and include the appropriate
units. [OH−] = SubmitMy AnswersGive Up Part B Find the pH of a
0.45 M aniline (C6H5NH2) solution. Express your answer using two
decimal places.

Calculate the pH at 25°C of a 0.97M solution of
anilinium chloride C6H5NH3Cl. Note that aniline C6H5NH2 is a weak
base with a pKb of 4.87. Round your
answer to 1 decimal place.

Consider a solution that contains both C6H5NH2 and C6H5NH3+.
Calculate the ratio [C6H5NH2]/[C6H5NH3+] if the solution has the
following pH values. (Assume that the solution is at 25°C.) (a) pH
= 4.69 (b) pH = 5.19 (c) pH = 5.03 (d) pH = 5.44

Calculate the pH of a solution that is 3:1 mixture by volume of
0.25M benzoic acid and 0.50M sodium benzoate respectively. The Ka
for benzoic acid is 6.28E-5. If the solution in a) was diluted 20x
with neutral water, what would be the resulting pH?

PLEASE SHOW WORK
4. Calculate the [H+], pH and percent dissociation of a solution
that is 0.50 M in HOCl and 0.60 M in NaOCl.
5. Calculate the [H+], pH and percent dissociation of a solution
that is 1.0 M in aniline (C6H5NH2) and 1.2 M in C6H5NH3Cl.
6. A solution contains 0.60 mol of NH4Cl and 0.30 mol of NH3 in
1.5 L. What is the pH if 0.070 mol of HCl is added to this
buffer?

Calculate the pH of the solution resulting from the addition of
85.0 mL of 0.35 M HCl to 30.0 mL of 0.40 M aniline
(C6H5NH2). Kb
(C6H5NH2) = 3.8 x
10-10
9.09
4.64
4.19
0.81
1.75
Please fully explain your answer

What is the total number of atoms in 4.65 g of aniline
(C6H5NH2)?

A 0.198 M weak acid solution has a pH of 4.12. Find
Ka for the acid.
Express your answer using two significant
figures.
Find the [OH−] of a 0.44 M aniline (C6H5NH2)
solution. (The value of Kb for aniline (C6H5NH2) is
3.9×10−10.)
pH=9.12

Determine the pH of the solution when 20.00 mL of 0.1563 M
aniline hydrochloride (C6H5NH3+Cl-) is titrated with 15.00 mL of
0.1249 M NaOH. (Ka(aniline)=2.40 x 10-5)

Calculate the Ka, for an unknown acid HX, given that a 0.25M
salt solution NaX has a pH of 8.48
answer: 2.7x10-4

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