Question

53. What is the pH of a 0.4283 M aqueous solution of ammonia?
K_{b} (NH_{3}) = 1.8x10^{-5}

Answer #1

Lets write the dissociation equation of NH3

NH3 +H2O -----> NH4+ + OH-

0.4283 0 0

0.4283-x x x

Kb = [NH4+][OH-]/[NH3]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.8*10^-5)*0.4283) = 2.777*10^-3

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

so, x = 2.777*10^-3 M

so.[OH-] = x = 2.777*10^-3 M

we have below equation to be used:

pOH = -log [OH-]

= -log (2.777*10^-3)

= 2.55

we have below equation to be used:

PH = 14 - pOH

= 14 - 2.56

= 11.44

Answer: 11.44

calculate the ph of .50 M NH4CL solution for NH3 , Kb= 1.8x10^-5
what is the net ionic equation for the reaction

calculate the pH of the following solution: 0.10 M NH3 with
KB=1.8x10^-5

1)
Kb for NH3 is 1.8x10^-5. What is the pOH of a .20M aqueous solution
of NH4Cl at 25 degrees Celsius? (Please show work)
2) Determine the pH of a 0.15 M aqueous solution of KF. For
hydrofluoric acid, Ka=7.0x10^-4. (Please show work)
3) An aqueous solution of _____ will produce a neutral
solution.
A) NaNO2
B) LiNO3
C) KF
D) Rb2CO3
E) NH4NO3

-What concentration of ammonia is required to have a solution
with a pH of 11.36? Kb = 1.8x10-5
-What is the pH of a 0.014M solution of NaF? Ka for
HF = 6.8 x 10-4
-What is the other product of the aqueous base dissociation
reaction of ethyl amine?
C2H5NH2(aq) + H2O(l)
⇌ ________ + OH-(aq)

Calculate the [OH-] pH and percent ionization for a
0.2 M aqueous solution of NH3. Kb = 1.8
x10-5
Thank you!

What is the pH of 0.30 M ammonia solution? Kb of Ammonia is 1.8
x 10 NH3 (aq) + H20 (l) NH; (aq) + OH-(aq)

Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5.What is
the pH of a 0.335 M ammonia solution?What is the percent ionization
of ammonia at this concentration?

Find the pH of a 0.218 M NH4Cl solution given that the Kb of NH3
= 1.8x10-5 at 25 °C.
The answer is 4.96 but I'm not sure how to get there.

What is the pH of a solution made by adding 1.451 moles of
ammonia (NH3, Kb = 1.8 x 10-5) to 537 mL of distilled water?

a. What is the pH of a buffer solution that is 0.24 M
NH3 and 0.24 NH4+? Kb
for NH3 is 1.8x10-5.
b. What is the pH if 17 mL of
0.27 M hydrochloric acid is added to
515 mL of this buffer?

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