Question

A 25.00 mL sample of a 0.250 M aqueous solution CH₃NH₂ (a weak base) is titrated with an 0.100 M aqueous solution of HCl (a strong acid.) The molecular and net ionic equation for the reaction is provided below. The Kb value used for CH3NH2 is 4.4x10^-4.

Find the pH of the solution after addition of 15.00 mL of the aqueous solution of HCl

Molecular: CH₃NH₂ (aq) + HCl (aq) → CH₃NH₃⁺ + Cl^—

Net ionic:         CH₃NH₂ (aq) + H⁺  →   CH₃NH₃⁺

Answer 1

And) This is an titration of weak e This is an example of base with strong Acid Cucak base =CH₂NH2 25.00ml 0.250M - CH3NH2 Strong Strong Acid -14 cl A 0.100M Strong acid added - 15 mL Amount of base (Cty Nitz) present in the solution. - 250.250 = 6.25 millimoles of C₂NH2 Amount of Acid (HCl) added = 0.100 x 15 -1.5millimoles of Amount of salt Formed from HO TEHNING TH Nigel - Amount of Hel added - 1.5 millimoles of Balt] Acid and base duart to give salt. The solution with salt and sare duesponsible for the Por OF Solution. Amount of base left in the solution COM₂Nt,] = [base] = 6.25 -1.5 -4.75 Millius esor (+ 3NH2 Kb=4.4x10-4 Pkb = -dog (kb) Pkb= 3.357 Pole = Pkb + dog [salt] [Base] - POH = 3.357 + dog [ 1.5] [4.75] - POH = 2.8 56 => PH+ Polt = 14 =) PH = 14- Polt = 14-2856 PH = 11.14 - Ansuur = 3.357 - 0.5006

Since volume of the solution is the same for both [Cr₂Nt ] and [ergaryel], the ratio of their amounts in millimoles can be taken as the ratio of their molar Concentrations.