Question

QUESTION 10 Calculate the pH during the titration of 30.00 mL of 0.1000 M methylamine, (CH3)NH2(aq), with 0.1000 M HCl(aq) after 12 mL of the acid have been added. Kb of methylamine - 3.6 x 10-4

Answer 1

The main concept of this question is to identify that the solution formed is a basic buffer solution.

Milimoles of CH₂ N M2 = 3080 1 2 3 Milimoles of HU = 12x0;l = 1.2 The required reaction : - (H₂NH₂ + Hill Che NH ₂ + c - → The solution of is a buffer solution. New molarity of CH₃NH₂> Total moles . Volume of solution - 1.8 X10" - 3 M 42 x 10-3 70 New molarity of CH₂ NH3t = 1.2x10 3 - 2M CH3NH₂ + tho. Kb Hz NH3t tour" 70 ir 42 x 103 Kbz (2/0 * 7 (x) : ks ~ 2 (x)) (36*). - [ou] = 3 x 3,6x16" 5.4 x 109 M Ou 5.48104 PMS - log MT= -log(10-1) = 10 +8.732 lol 54 10.73

Once we are able to found that,we can also directly use the formula for basic buffer solution.  The formulae is

pOH = pKB + log([acid]/[base]) , pH=14 - pOH