Question

Consider the titration of a 23.0 −mL sample of 0.170 M CH3NH2 with 0.155 M HBr. Determine each of the following.

C) the pH at 6.0 mL of added acid. Express your answer using two decimal places.

D) the pH at one-half of the equivalence point. Express your answer using two decimal places.

E) the pH at the equivalence point .Express your answer using two decimal places.

F) the pH after adding 5.0 mL of acid beyond the equivalence point. Express your answer using two decimal places.

Answer 1

Answer- We are given, 23.0 mL of 0.170 M CH₃NH₂, [HBr] = 0.155 M

C) the pH at 6.0 mL of added acid –

moles of CH₃NH₂= 0.170 M * 0.023 L = 0.00391 mole

moles of HBr = 0.155 M * 0.006 L = 0.00093 moles

CH₃NH₂ + HBr -----> CH₃NH₃⁺ + Br^-

0.00391 0.00093      0.00093

New moles of CH₃NH₂ = 0.00391 - 0.00093 = 0.00298 moles

Moles of CH₃NH₃⁺ = 0.00093 moles

New molarity calculation, so total volume = 23 +6 = 29 mL

[CH₃NH₂] = 0.00298 moles / 0.029 L

              = 0.103 M

[CH₃NH₃⁺ ] = 0.00093 mole / 0.029 L = 0.0321 M

We know, pKb for CH₃NH₂ = 3.36

pOH = pKb + log [CH₃NH₂⁺] / [CH₃NH₂]

         = 3.36 + log (0.0321 / 0.103)

         = 2.85 + (-0.505)

         = 2.85

pH = 14- pOH

       = 14 – 2.85

        = 11.14

D) the pH at one-half of the equivalence point

We know at one-half of the equivalence point the base and it conjugate acid have same molarity then there is

pOH = pKb

so, pOH = 3.36

pH = 14- pOH

       = 14 – 3.36

        = 10.64

E) the pH at the equivalence point –

We know at the equivalence point. Moles of acid and base are equal, so all base reacted with HBr and form the conjugate acid –

CH₃NH₂ + HBr -----> CH₃NH₃⁺ + Br^-

0.00391 0.00391       0.00391

Moles of CH₃NH₃⁺ = 0.00391 moles

We need to calculate volume of HBr

We know moles of HBr and molarity

So, volume of HBr = 0.00391 moles / 0.155 M

                               = 0.0252 L

                                = 25.2 mL

New molarity calculation, so total volume = 25 +25.2 = 50.2 mL

[CH₃NH₃⁺] = 0.00391 mole / 0.0502 L = 0.0779 M

We need to put ICE chart

      CH₃NH₃⁺ + H₂O ------> CH₃NH₂ + H₃O⁺

I     0.0779                            0              0

C    -x                                 +x             +x

E 0.0779-x                         + x            +x

Ka = 1*10^-14 / 4.38*10^-4

        = 2.88*10^-11

Ka = [CH₃NH₂] [H₃O⁺] / [CH₃NH₃⁺]

2.88*10^-11 = x *x / 0.0779 –x

x² = 2.88*10^-11 * 0.0779

x = 1.33*10^-6 M

[H₃O⁺] = x= 1.33*10^-6 M

pH = - log [H₃O⁺]

     = - log 1.33*10^-6 M

    = 5.87

F) the pH after adding 5.0 mL of acid beyond the equivalence point-

We know at equivalence point we need 25.2 mL and we added more 5.0 mL

So, total volume = 25+30.2 mL = 55.2 mL

Moels of HBr = 0.155 *0.0302 L

                       = 0.00468 moles

[CH₃NH₃⁺] = 0.00468 mole / 0.0552 L = 0.0848 M

We need to put ICE chart

      CH₃NH₃⁺ + H₂O ------> CH₃NH₂ + H₃O⁺

I     0.0848                             0              0

C    -x                                   +x             +x

E 0.0848-x                             + x            +x

Ka = 1*10^-14 / 4.38*10^-4

         = 2.88*10^-11

Ka = [CH₃NH₂] [H₃O⁺] / [CH₃NH₃⁺]

2.88*10^-11 = x *x / 0.0848 –x

x² = 2.88*10^-11 * 0.0848

x = 1.39*10^-6 M

[H₃O⁺] = x= 1.39*10^-6 M

pH = - log [H₃O⁺]

     = - log 1.39*10^-6 M

    = 5.86