Question

6. Sketch the titration curve and calculate the pH of the equivalence point for the titration of 30ml of 0.20M CH3NH2 with 0.10M HBr.

Answer 1

Ginut 30 mL of 0,20m CH₂ NHL (ko = 5,0 x 10" 0.10 M HBV > они L x0,20 м mana cң, мн, а = 6 mmol mous a oilo At equivalence point on mohs of the Nth = moles of HBr = 6mmo Promote 6 mmol so volume of HBr = molarity volume of her a 60 ml Total mume & 30mL + 60 mL = 90mL CANH + HBr - Ch Nht + br- 6 mmol o o 6mmol -€ Е + te с 6 mmol 6 mann 0 о CS Scated with CamScanner

"CH NHT = 6 mmol Гd94 | им с бо м с [CHANH t = 6 mmol = 0.1m 60 м tentujale quid) res erver Now i +ho — сиң, Nң, +ң от ЛЕНА си, Nң 01 – x о 1-х I с Е + и 1.0x10ls соң, мн.) Сң о*) Со ни, so xlog х. 2-охо - o | - о- о| x 10 = — 1 4 2. 2. б хоч - [Hoot] = x = 1.4 x 10 6 CS Canned with pHC - log [ hy of) = -log 1.4x1016

pH = 5.85 , at equivalence pont 5.85 Equivalence point VyBn - 60 mL - Scanned with

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