Question

The Kb for methylamine, CH3NH2, at 25 C is 4.4 x 10-4 A. Write the chemical equation for the equilibrium that corresponds to Kb. B. By using the value of Kb, calculate ΔGo for the equilibrium in part A. C. What is the value of ΔG at equilibrium? D. What is the value of ΔG when [H+] = 1.6 ×10-8 M, [CH3NH3+] = 5.8 ×10-4 M, and [CH3NH2] = 0.130 M?

Answer #1

ΔG˚ = 19.1 kJ Kb for (CH3NH2) at 25˚C is
4.4x10-4
What is the value of ΔG when [H+] = 1.6
×10-8 M, [CH3NH3+] =
5.1x10-4, and [CH3NH2] = 0.130
M?
I got the answer as 49.1 kJ using the following equation:
ΔG= -RT ln Kb
ΔG= (-8.314 J/mol*K)(298.15 K)(ln(2.4519x10-9) =
49,146 J/mol = 49.1 kJ/mol
The answer wants it in kJ not kJ/mol. What am I doing wrong?

Calculate the pH of a 0.50 M solution of
methylamine(CH3NH2, Kb = 4.4 x
10-4.)
a.
1.83
b.
10.64
c.
12.17
d.
5.47
e.
8.53

Part A
Find [OH−] of a 0.33 M methylamine (CH3NH2) solution.
(Methylamine has a Kb value of 4.4×10−4.)
Express the concentration in moles per liter to two significant
figures.
Part B
Find the pH of the 0.33 M methylamine (CH3NH2)
solution.
Express the pH of the solution to two decimal places

The salt formed by the reaction of the weak base methylamine,
CH3NH2, with the strong acid nitric acid is methylammonium nitrate,
CH3NH3NO3. What is the pH of a 0.088 M solution of methylammonium
nitrate at 25∘C given that the value of Kb for methylamine is
4.4×10−4?

Calculate the pH of a 0.22 M CH3NH3Br solution. Kb(CH3NH2) = 4.4
×10–4

A 102.0 mL sample of 0.100 M methylamine (CH3NH2;Kb=3.7×10−4) is
titrated with 0.270 M HNO3. Calculate the pH after the addition of
each of the following volumes of acid:
Part A: 0.0 mL
Part B: 18.9 mL
Part C: 37.8 mL
Part D: 56.7 mL

A 111.8 mL sample of 0.100 M methylamine
(CH3NH2;Kb=3.7×10−4) is titrated with 0.245
M HNO3. Calculate the pH after the addition of
each of the following volumes of acid.
Part A
0.0 mL
Part B
22.8 mL
Part C
45.6 mL
Part D
68.4 mL

HNO2 ( Ka=4.5 x 10^-4)
HCN (Ka= 4.9 x 10^-10)
CH3NH2 (Kb=4.4 x 10^-4)
HONH2 (Kb=1.1 x 10^-8)
Use this data to rank the following solutions in order of
increasing pH. In other words, select a '1' next to the solution
that will have the lowest pH, a '2' next to the solution that will
have the next lowest pH, and so on..
RANK THESE ONES:
0.1 M NaBr
0.1 M CH3NH3Cl
0.1 M KNO2
0.1 M HONH3Br

Write the acid form of the four equations describing a 10^-2
molar solution of methylamine (CH3NH2, kb = 5.25 x 10^-4). DO NOT
SOLVE. What is the pH of this solution?

In the titration of 40.00 mL of 0.500 M methyl amine, CH3NH2 (Kb
= 4.4 x 10–4) with 0.200 M HCl, calculate the pH of the resulting
solution when the titration is 1/4 of the way to the equivalence
point.

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