Question

9. Calculate the pH for each of the following cases in the titration of 35.0 mL of 0.200-M methylamine, CH3NH2(aq), with 0.200 M HCl(aq): (a) before addition of any HCl(aq) (b) after addition of 17.5 mL of HCl(aq) (c) after addition of 34.9 mL of HCl(aq) (d) after addition of 35.0 mL of HCl(aq) (e) after addition of 35.1 mL of HCl(aq)

Answer 1

a) pH = 11.97

b) pH = 10.66

c) pH = 7.96

d) pH = 5.83

e) pH = 3.7

Step-by-step explanation:

a) The reaction is

CH₃NH₂ + H₂O = CH₃NH₃+ + OH-

`K_(b) =([CH_(3)NH_(3)]+ [OH-] )/([CH_(3)NH_(2) ] )\\4.6x10^(-4) =(x*x)/(0.2-x) \\x=0.00936 M`

`pOH=-log(0.00936)=2.03\\pH=14-2.03=11.97`

b)

`[acid]=(M_(2)V_(2) )/(V_(1)+V_(2) ) =(0.2*17.5)/(35+17.5) =0.0667M\\[base]=(M_(1)V_(1) )/(V_(1)+V_(2) ) =(0.2*35)/(35+17.5) =0.13M`

base left = 0.13 - 0.0667 = 0.066 M

`pOH=pK_(b) +log(([salt])/([base]) )=3.34+log(0.0667/0.066)=3.34\\pH=14-3.34=10.66`

c)

`[acid]=(0.2*34.9)/(35+34.9) =0.0998M\\\\ [base]=(0.2*35)/(35+34.9) =0.1M\\base-left=0.1-0.0998=0.0002M\\pOH=3.34+log(0.0998/0.0002)=6.04\\pH=14-6.04=7.96`

d) at equivalence point:

`[acid]=(0.2*35)/(35+35) =0.1M\\[base]=(0.2*35)/(35+35) =0.1M\\\\pH=7-(1)/(2) (pK_(b)+ log(c))=7-(1)/(2)(3.34+log(0.1))=5.83 \\`

e)

`[acid]=(0.2*35.1)/(35+35.1) =0.1M\\[base]=(0.2*35)/(35+35.1) =0.0998M\\acid-left=0.1-0.0998=0.0002M\\pH=-log(0.0002)=3.7\\`

Answer 2

a) 13.3

b) 11.8

c) 10.5

d) 7

e) 3.5

Step-by-step explanation:

The reaction for the titration is given by :

HCl + CH3NH2 ---> CH3NH3Cl

1. 1. 1

Given the CB = 0.2M, VB = 35mL

CA = 0.2M,

Where C represent the concentration of the reactants and V is the volumes and B is the base and A is the acid.

Let n represent the moles

n = concentration × volumes

n(B) = 0.2 × (35/1000) = 0.007 mol

(a) so before addition of HCl, the solution will be alkaline.

poH= -log[CH3NH2] = -log[0.2]

pOH= 0.699

pH + pOH = 14

pH = 14-0.699

pH = 13.30

pH = 13.3

(b)After that addition of 17.5ml:

The reaction ratio is 1:1

n(A) = 0.2 × (17.5/1000)

n(A) = 0.0035 mol.

From the reaction:

0.0035 mol of the base will require 0.0035 mole of the acid. So the base(CH3NH2) is in excess and so determine the pH

n(B) excess = 0.007-0.0035

= 0.0035mol

The concentration of the excess base in solution is given by

[CH3NH2] =0.0035/(0.035+0.0175)

The volume becomes the total volume of solution.

[CH3NH2] = 0.0667

pOH = -log[CH3NH2]

pOH = -log[0.0667]

pOH = 2.18

pH = 14-2.18

pH = 13.82

pH = 11.82

pH = 11.8

(c) After the addition of 34.9mL of acid

n(A) = 0.2 × (34.9/1000)

= 0.2 × 0.0349

= 0.00698 mol

So compare with base, the base is in excess by

n(B) excess = 0.007-0.00698

= 0.00002 mol

[CH3NH2] excess = 0.00002/(0.035 + 0.0349)

= 0.000286

pOH = log[0.000286]

pOH = 3.54

pH = 14 -3.54

pH = 10.46

pH = 10.5

(d)After the addition of 35mL

n(A) = 0.2 × (35/1000)

= 0.2 × 0.0035

= 0.007

At this point the number of mole of the acid is equal to the mole of the base; no reactants is in excess, So the solution is neither basic nor acidic, that is neutral. the pH here is assumed 7.

e) After the addition of 35.1mL of acid

n(A) = 0.2 × 0.0351

= 0.00702

Comparing the acid to the base in 1:1, the acid is in excess by

0.00702-0.007 = 0.00002.

Hence the concentration of H+ in solution is given by:

[H+] = 0.00002/(0.035+0.0351)

= 0.000285

pH = -log[0.000285]

pH = 3.545

pH ~= 3.5