Question
Asked Dec 2, 2019
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Calculate the pH at the equivalence point in the titration of 45.0 mL of 0.170 M methylamine

(
Kb = 4.4
×
10−4
)

with 0.275 M HCl.

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Expert Answer

Step 1

Given,

Volume of methylamine = 0.170 M = 0.17 mol/L

Molarity of methylamine = 45.0 mL = 0.045 L     (1 mL = 0.001 L)

Molarity of HCl = 0.275 M = 0.275 mol/L

Moles of methylamine can be calculated as :

Moles
Molarity olume in litre
Moles Molarity x Volume in litre
Moles 0.17 mol/L x 0.045 L
Moles 0.00765 mol
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Moles Molarity olume in litre Moles Molarity x Volume in litre Moles 0.17 mol/L x 0.045 L Moles 0.00765 mol

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Step 2

The reaction of methylamine with strong acid, HCl can be written as :

 CH3NH2 (aq) + H+ (aq) → CH3NH3+ (aq)

At equivalence point, the moles of methyl amine will be equal to the moles of HCl. Thus, we can calculate the volume of HCl used as:

At equivalence point
Moles of HCl moles of methylamine 0.00765 mol
Moles
Molarity
Volume
Moles
Volume
Molarity
0.00765 mol
Volume
0.275 mol/L
Volume
0.028 L
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At equivalence point Moles of HCl moles of methylamine 0.00765 mol Moles Molarity Volume Moles Volume Molarity 0.00765 mol Volume 0.275 mol/L Volume 0.028 L

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Step 3

At equivalence point, the concentration of CH3NH3+ ...

Total volume = Volume of HCl
Volume of CH3NH2
Total volume
0.028 L 0.045 L = 0.073 L
Moles of CH3NH3 = 0.00765 mol
0.00765 mol
Molarity of CH3NH3
= 0.105 M
0.073 L
help_outline

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Total volume = Volume of HCl Volume of CH3NH2 Total volume 0.028 L 0.045 L = 0.073 L Moles of CH3NH3 = 0.00765 mol 0.00765 mol Molarity of CH3NH3 = 0.105 M 0.073 L

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