Question

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

Answer #1

The net ionic equation is:

NH4+ (aq) —> NH3 (aq) + H+ (aq)

we have below equation to be used:

Ka = Kw/Kb

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

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

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

Ka = 5.556*10^-10

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

0.5 0 0

0.5-x x x

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

Ka = 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

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.556*10^-10)*0.5) = 1.667*10^-5

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

so, x = 1.667*10^-5 M

So, [H+] = x = 1.667*10^-5 M

we have below equation to be used:

pH = -log [H+]

= -log (1.667*10^-5)

= 4.78

Answer: 4.78

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

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.

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

53. What is the pH of a 0.4283 M aqueous solution of ammonia?
Kb (NH3) = 1.8x10-5

What is the pH of a 0.422 M Ammonium Chloride, NH4Cl,
solution if the Kb for ammonia, NH3, is 1.8 x
10-5?

Calculate the pH of a 1 L solution made by mixing 0.25mol of NH3
and 0.35mol NH4Cl. (Kb = 1.8x10^- 5). What is the new pH if 0.15
mol HCl is added to the solution?

a)You obtain a 0.817 M solution of NH4Cl. Knowing that the Kb of
NH3 is 1.8 * 10-5, what is the Ka of NH4+?
b) What is the [H+] in
the solution in part a?
c) What is the pH of the solution in part
a?

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?

NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts
as a weak acid. What is the pH of a solution that is 0.059 M in
NH4Cl at 25 °C?
pH =

A buffer consists of 0.26 M NH4Cl and 0.36 M NH3. What is the pH
of the buffer after 0.003 moles of Ca(OH)2 are added to 0.10 L of
this buffer solution. Kb for NH3 is 1.8×10‒5.

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 5 minutes ago

asked 6 minutes ago

asked 12 minutes ago

asked 12 minutes ago

asked 23 minutes ago

asked 23 minutes ago

asked 26 minutes ago

asked 27 minutes ago

asked 32 minutes ago

asked 32 minutes ago

asked 38 minutes ago

asked 47 minutes ago