Question

Consider the titration of a 25.0 mL sample of 0.170 molL−1 CH3NH2 (Kb=4.4×10−4) with 0.150 molL−1 HBr. Determine each quantity:

Part A

the volume of added acid required to reach the equivalence point

Part B

the pH at 6.0 mL of added acid

Express your answer using two decimal places.

Part C

the pH at the equivalence point

Express your answer using two decimal places.

Part D

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

Express your answer using two decimal places.

Answer 1

Part-A:

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

Initial moles of CH₃NH₂ = M*V = 0.170 mol/L * 25.0 mL * (1L / 1000 mL) = 0.00425 mol

Since mole ratio is 1:1,

moles of HBr required to reach equivalence point = Initial moles of CH₃NH₂ =  0.00425 mol

Given [HBr] = 0.150 mol/L

=> Volume of HBr required = 0.00425 mol * (1L / 0.150 mole) = 0.02833 L = 28.3 mL (Answer)

PArt-B:

Moles of acid added = M*V = 0.150 mol/L * 6.0 mL * (1L / 1000 mL) = 0.0009 mol

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

I: 0.00425, 0.0009 ------ 0 ------------------ 0

C: - 0.0009, - 0.0009 --- + 0.0009, ------ + 0.0009

E: 0.00335, 0 mol ------ 0.0009 mol --- 0.0009 mol

Since the solution contains both CH₃NH₂ and CH₃NH₃⁺(aq), it acts as buffer solution.

Hence the pH can be found from Hendersen equation:

pOH = pKb + log (CH₃NH₃⁺(aq) mol) / ( CH₃NH₂ mol)

=> pOH = - log(4.4*10^-4) + log(0.0009 / 0.00335)

=> pOH = 2.786

=> pH = 14 - 2.786 = 11.21 (Answer)

PArt-C:

To reach the equivalence point we have add 28.3 mL HBr

Hence total volume at equivalence point, Vt = 25.0 mL + 28.3 mL = 53.3 mL = 0.0533 L

Also at equivalence point all of the CH₃NH₂ are converted to CH₃NH₃⁺(aq).

Hence moles of CH₃NH₃⁺(aq) = 0.00425 mol

=> [CH₃NH₃⁺(aq)] =  0.00425 mol / Vt = 0.00425 mol / 0.0533 L = 0.07974 mol/L

At equivalence point, CH₃NH₃⁺(aq) undergoes dissociation to form CH₃NH₂ and H₃O⁺

--CH₃NH₃⁺(aq) -----> CH₃NH₂ + H₃O⁺ : Ka = Kw / Kb = 10^-14 / 4.4*10^-4 = 2.27*10^-11  

I:  0.07974 M ----------- 0 ---------- 0

C: - X --------------------- +X -------- +X

E: (0.07974-X), --------- X ---------- X

Ka = 2.27*10^-11​ = X² / (0.07974-X)

=> X = [H₃O⁺] = 1.346*10^-6

=> pH = - log(1.346*10^-6 ) = 5.87 (Answer)

Part-D:

Total volume, Vt = 53.3 mL+ 6 mL = 59.3 mL = 0.0593 L

Moles of H₃O⁺ (or HBr) added = M*V = 0.150 * 0.006 = 0.0009 mol

=> [H₃O⁺] = 0.0009 mol / Vt = 0.0009 mol / 0.0593 L = 0.015177 M

=> pH = - log(0.015177) = 1.82 (Answer)