Question

#27 Weak Base Buffer 1. a. Write the equation for the dissociation of the weak base methylamine, CH3NH2. CH3NH2 + H2O ? HC3NH2+ + OH- WB CA b. What is the pH of a 1.556M solution of methylamine? kb = 4.4x10^-4 CH3NH2 + H2O ? HC3NH2+ + OH- I 1.556 0 0 C -x +x +x E 1.556-x x x 4.4x10^-4 = x^2 / (1.556-x) 4.4x10^-4 - 4.4x10^-4x = x^2 4.4x10^-4x - 6.8x10^-4 = x^2 x = .026 pOH = -log(OH-) pOH = -log(.026) pOH = 1.59 pH = 14-1.59 = 12.41 c. What is the pH of a buffer made by mixing 1.556 mol methylamine with 1.556 mol methylammonium chloride (CH3NH3+). CH3NH2 + H2O ? HC3NH2+ + OH- I C E d. What is the pH of this buffer after the addition of 0.0056mol HCl?

          #27 Weak Base Buffer
1. a. Write the equation for the dissociation of the weak base methylamine, CH3NH2.
CH3NH2 + H2O ? HC3NH2+ + OH-
WB CA
b. What is the pH of a 1.556M solution of methylamine? kb = 4.4x10^-4
CH3NH2 + H2O ? HC3NH2+ + OH-
I 1.556 0 0
C -x +x +x
E 1.556-x x x
4.4x10^-4 = x^2 / (1.556-x)
4.4x10^-4 - 4.4x10^-4x = x^2
4.4x10^-4x - 6.8x10^-4 = x^2
x = .026
pOH = -log(OH-)
pOH = -log(.026)
pOH = 1.59
pH = 14-1.59 = 12.41
c. What is the pH of a buffer made by mixing 1.556 mol methylamine with 1.556 mol methylammonium chloride (CH3NH3+).
CH3NH2 + H2O ? HC3NH2+ + OH-
I
C
E
d. What is the pH of this buffer after the addition of 0.0056mol HCl?

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#27 Weak Base Buffer
1. a. Write the equation for the dissociation of the weak base methylamine, CH3NH2.
CH3NH2 + H2O ? HC3NH2+ + OH-
WB CA
b. What is the pH of a 1.556M solution of methylamine? kb = 4.4x10^-4
CH3NH2 + H2O ? HC3NH2+ + OH-
I 1.556 0 0
C -x +x +x
E 1.556-x x x
4.4x10^-4 = x^2 / (1.556-x)
4.4x10^-4 - 4.4x10^-4x = x^2
4.4x10^-4x - 6.8x10^-4 = x^2
x = .026
pOH = -log(OH-)
pOH = -log(.026)
pOH = 1.59
pH = 14-1.59 = 12.41
c. What is the pH of a buffer made by mixing 1.556 mol methylamine with 1.556 mol methylammonium chloride (CH3NH3+).
CH3NH2 + H2O ? HC3NH2+ + OH-
I
C
E
d. What is the pH of this buffer after the addition of 0.0056mol HCl?

Added by Amparo C.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Madhur L

08/02/2023

Step 1

556 M solution of methylamine. We are given the Kb value as 4.4 x 10^-4. We can set up an ICE table: $$\begin{array}{c|c|c|c} & CH_3NH_2 & + & OH^- \\ \hline I & 1.556 & & 0 \\ C & -x & & +x \\ E & 1.556-x & & x \end{array}$$ Now, we can write the expression for Show more…

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#27 Weak Base Buffer 1. a. Write the equation for the dissociation of the weak base methylamine, CH3NH2. CH3NH2 + H2O ⇌ CH3NH3+ + OH- b. What is the pH of a 1.556M solution of methylamine? kb = 4.4 x 10^-4 pOH = -log(0.026) = 1.59 pH = 14 - 1.59 = 12.41 c. What is the pH of a buffer made by mixing 1.556 mol methylamine with 1.556 mol methylammonium chloride (CH3NH3+). d. What is the pH of this buffer after the addition of 0.0056mol HCl?

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00:01 Hi, in this question we are having 4 parts.

00:02 In the first part of the question we are asked to write the dissociation equation of weak base methylamine.

00:08 We can say methylamine is ch3nh2 when it dissociate in water then we can say that the nitrogen is having lone pair of electron it will accept the h plus from h2o as a result of which we are getting ch3nh3 positive ion and there will be formation of oh minus ion.

00:39 This is the dissociation equation of weak base methylamine.

00:43 Now in second part of the question we are asked to calculate the ph of solution of methylamine.

00:54 We are given with the concentration of methylamine equals to 1 .556 molar.

01:04 Now we know that methylamine is a weak base.

01:07 We can calculate its oh minus ion concentration that will be equal to square root kb into c.

01:16 Kb is the base dissociation constant and c is the concentration.

01:20 We can substitute the values here in this manner on solving this we are getting the oh minus ion concentration equals to 0 .0261 molar.

01:32 Now from this oh minus ion concentration we can calculate the poh.

01:36 It will be equal to minus log oh minus ion concentration.

01:41 We can substitute the values here in this manner on solving we are getting the poh value as 1 .58.

01:50 We can calculate the ph as it will be equal to 14 minus poh.

01:57 We can substitute the values here in this manner on solving we are getting the ph of solution equals to 12 .42...

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