Use the following information to answer the next five questions. The following b

Question

Use the following information to answer the next five questions. The following buffer was prepared: 50. mL 2.0 M methylamine (CH_3HN_2) solutions and 50. mL of 2.0 M methylamine hydrochloride (CH_3NH_3Cl) solutions. The K_b for methylamine is 4.4 times 10^-4. What is the pH of the buffer solutions? 11.12 10.64 12.32 0.48 5.32 What is the pH of the resulting solution after adding 100 ml of 1.0 M HCl to the original buffer solution? 11.12 10.64 12.32 0.48 5.32 What is the pH of the resulting solution after adding 200 ml of 1.0 M HCL to the original buffer solution? 11.12 10.64 12.32 0.48 5.32 What is the pH of the resulting solution after adding 50 mL of 1.0 M KOH to the original buffer solution? 11.12 10.64 12.32 0.48 5.32

Explanation / Answer

SOLUTION:

Q40.

pH = pKa + log[CH3NH2] / [CH3NH3Cl]

Kb X Ka = Kw; 4.4 X 10-4 X Ka = 1.0 X 10-14 ; Ka =  1.0 X 10-14 / 4.4 X 10-4 = 2.3 X 10-11

pKa = - logKa = - log  2.3 X 10-11 = 10.64

Therefore pH = 10.6 + log[2] / [2] = 10.6 OPTION (B) IS CORRECT ANSWER.

Q41. Moles of HCl added = Molarity X volume = 1.0 X 100 = 100mmol

Moles of CH3NH2 = 2 X 50 = 100mmol

100mmol of HCl will react with 100mmol of CH3NH2 to form 100mmol of CH3NH3Cl as

CH3NH2 + HCl ------->CH3NH3Cl

Total CH3NH3Cl = 100mmol + 100mmol = 200mmol = 0.2 moles

The pH of solution will be regulated by CH3NH3Cl as:

CH3NH3+ <---------> CH3NH2 + H+

Concentration of [CH3NH3+ ] no. of moles / volume in liters = 0.2 / 0.2L = 1M ;

Bz total volume 100mL (buffer) + 100mL (HCl) = 200mL = 0.2L

Ka = [CH3NH2][H+ ] / [CH3NH3+ ] = x2 / 1 - x

x can be neglected in denomenator as Ka is very small.

2.3 X 10-11 = x2 ; x = 4.8 X 10-6M

pH = -log [4.8 X 10-6] = 5.2 OPTION (E) IS CORRECT ANSWER.

Q42. Total moles of HCl added = 200 X 1 = 200mmoles

Out of 200mmoles (or 0.2 mole) of HCl 100mmols (or 0.1 mole) will react with CH3NH2 and 100mmol (or 0.1mole) will remain unreacted. Hence pH will be controlled by HCl

Concentration of [HCl] = n / V = 0.1 / 0.3 = 0.33M

pH = - log[0.33] = 0.48 HENCE OPTION (D) IS COORECT ANSWER.

Q43. Moles of KOH added = 50 X 1.0 = 50mmol = 0.05 mole

moles of CH3NH3Cl in origial solution = 50 X 2 = 100mmol = 0.1 mole

0.05 mole of KOH will react with 0.05 mole of CH3NH3Cl as torm 0.05 mole of CH3NH2

CH3NH3Cl + KOH ---------> KCl + CH3NH2 + H2O

Hence moles of CH3NH3Cl that will remain unreacted = 0.1 - 0.05 = 0.05

moles of CH3NH2 = 0.1 + 0.05 = 0.15 mole

Conecentration of [CH3NH2] = n / V ; V = 100 + 50 = 150mL = 0.15L

[CH3NH2] = 0.15 / 0.15 = 1

[CH3NH3Cl] = 0.05 / 0.15 = 0.33

pH = pKa + log[CH3NH2] / [CH3NH3Cl]

pH = 10.64 + log[1] / [0.33] = 11.12 Hence option (A) is correct answer.

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