Question

I have had such a hard time figuring out this problem please show with detailed response! thank you

Consider the titration of a 23.0?mL sample of 0.175M CH3NH2 with 0.155M HBr. Determine each of the following.

1.the initial pH

2.the volume of added acid required to reach the equivalence point

3.the pH at 6.0mL of added acid

4.the pH at one-half of the equivalence point

5.the pH at the equivalence point

6.the pH after adding 6.0mL of acid beyond the equivalence point

Answer 1

Given : Volume and molarity of CH3NH2 : 23.0 mL and 0.175 M

Molarity of HBr= 0.155 M

1). Initial pH

We calculate the pH by using molarity and volume of base given. To get equilibrium concentration of OH- we set up equilibrium equation and use kb value of give base. Once we get OH- concentration then by using it we get its pOH and then pH.

                 Kb value of CH3NH2 = 4.2E-4

ICE :

        CH3NH2 (aq)   + H2O (l)    ------- > CH3NH3+(aq) + H3O+ (aq)

I       0.175 M                                           0                              0

C       -x                                                +x                            +x

E   (0.175- x)                                       x                                  x

Kb = 4.2E-4 = x² / (0.175-x)

4.2E-4 = x² / (0.175 -x)

Lets use 5 % approximation

4.2E-4 = x² / 0.175

x² = 0.0000735

x = 0.00857

Lets check 5% approximation

Percent ionization

= x / 0.175 * 100
= 4.9 %

That is less than 5% that suggests our approximation is valid.

x = [OH-] = 0.00857 M

pOH= -log [OH-] = -log (0.00857 )

= 2.067

pH = 14 – pOH = 14- 2.067 = 11.93

2). The volume of acid required to reach the equivalence point.

We know at equivalence point the mole of acid = moles of base.   ( In the reaction of CH3NH2 and HBr, mol ratio of HBr to CH3NH2 is 1:1 )

Lets calculate moles of base = volume in L * Molarity

= 0.023 L * 0.175 M

= 0.004025 mol

Mol of acid HBr = 0.004025 mol

Now we are given molarity of HBr

Volume in L = mol / molarity

= 0.004025 mol / 0.155 M

= 0.02597 L

Volume in mL

= 0.02597 L * 1000 mL /1 L

= 26.0 mL

3).

The pH when 6.0 mL HBr is added

Mols of HBr

= 0.006 L * 0.155 M

= 0.00093 mol HBr

We know HBr reacts with CH3NH2 as follow

CH3NH2 (aq) + HBr (aq) ---- > CH3NH3+ (aq)   + Br- (aq)

From the above reaction we get that one mol of base is reduced and 1 mol of conjugate acid is

Formed.

Henderson equation

pOH = pkb + log ( [conj acid] / [base])

we know volume is same for all the species, so we just put the value of moles.

Pkb = -log kb = - log (4.2E-4) = 3.77

pOH = pkb + log ( [conj acid] / [base])

pOH = 3.377 + log [(0.00093 mol / ( 0.004025 – 0.00093 )]

= 2.85

pH = 14 -2.85 = 11.14

4). pH at half equivalence point

At half equivalence pOH = pkb

= 3.377

5) pH at equivalence point.

Lets write the reaction again.

CH3NH2 (aq) + HBr (aq) ---- > CH3NH3+ (aq)   + Br- (aq)

0.004025           0.004025             0

-0.004025       - 0.004025             0.004025

0                      0                      0

At equivalence point there remains moles of CH3NH3+ (aq) = 0.004025 mol

Molarity of CH3NH3(+) aq = 0.004025 mol / (0.023 L + 0.02597 L )

= 0.0822 M

ICE

          CH3NH3 + (aq) + H2O (l)    ---- > CH3NH2 (aq)   + H3O+ (aq)

I              0.0822                                              0                      0

C          -x                                                         +x                    +x

E          (0.0822 –x)                                          x                      x

Ka = 1.0E-14/ kb = 1.0E-14 / 4.2E-2 = 2.38E-11 = x²/ ( 0.0822-x)

Since value of ka is very small we neglect x in the denominator.

2.38E-11 = x²/ 0.0822

x = 1.402E-06

x = [H3O+] = 1.40E-6 M

pH = -log [H3O+] = -log ( 1.40E-6) = 5.85

6).

The pH after addition of 6.0 mL acid

Number of moles of acid = 0.006 L * 0.155 M

= 0.00093 mol

Final volume of acid = 0.02597 L + 0.006 L

= 0.03197 L

Total volume = volume of acid + volume of base

= 0.03197 L + 0.023 L

= 0.05497 L

HBr is monoprotic strong acid and its molarity = [H3O+]

Molarity of HBr = 0.00093 mol / 0.05497 L

= 0.017 M

pH = -log ( 0.017)

= 1.77