Consider the titration of a 25.0 mL sample of 0.170 M CH3NH2 with 0.145 M HBr. D

Question

Consider the titration of a 25.0 mL sample of 0.170 M CH3NH2 with 0.145 M HBr. Determine each of the following.

he initial pH

please explain in detail!

Express your answer using two decimal places.

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Part B

the volume of added acid required to reach the equivalence point

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Part C

the pH at 5.0 mL of added acid

Express your answer using two decimal places.

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Part D

the pH at one-half of the equivalence point

Express your answer using two decimal places.

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Part E

the pH at the equivalence point

Express your answer using two decimal places.

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Part F

the pH after adding 5.0 mL of acid beyond the equivalence point

Express your answer using two decimal places.

Explanation / Answer

We consider the initial concentration of OH- ions as x

Let us make an I.C.E. table for ease:



I...........0.170............0.........0
C...........-x...............+x.......+x
E........0.170-x..........x..........x

Kb = x2 / (0.170-x)

4.4 * 10-4 = x2 / (0.170-x)
Thus on solving we get: x = 8.65*10-3

i .e.[OH-] = 8.65*10-3 )
pOH = - log(8.65*10-3 ) = 2.06
pH = 14 - pOH = 14 – 2.06
pH = 11.94
Therefore initial pH = 11.94

b) The equivalence point is reached when moles of acid equals moles of base.

M1V1 = M2V2
(0.170)(25) = (0.145)(V2)
V2 = 29.31 ml

Thus volume of acid required to reach equivalence point = 29.3ml

c) When you add 5mL of acid, the volume changes.
Total volume = 29.3ml + 5ml = 34.3ml

Initial mol CH3NH2
0.170 mol CH3NH2 / L * 0.025 L = 0.00425 mol CH3NH2

Added mol HBr
0.145 mol HBr / L * 0.005L = 0.000725mol HBr = [acid]

The added

This will give 0.000725 moles of CH3NH3+

Moles CH3NH2 unreacted = (0.00425 - 0.000725) = 0.003525 = [base]

Determine the Ka value
Kw = Ka * Kb
Ka = (1.0*10-14) / (4.4 *10-4) = 2.27*10-11

pH = pKa + log ([base]/[acid])
pH = -log(2.27*10-11) + log [(0.003525)/(0.000725)]
pH = 10.64 + 0.68
pH = 11.32
pH on addition of 5ml HBr = 11.32

d) Half of the equivalence point.
The mol ratio of base and acid will be equal at ½ of equivalence point.
Therefore [Base] / [acid] = 1:1

pH = pKa + log ([base]/[acid])
pH = 10.64 + log (1/1)
pH = 10.64
Therefore pH at half of the equivalence point = 10.64

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