Question

7. Identify the titrant in each pair that wou ld provide a greater pH change about the equivalence point in the titration of a 0.100 M HCI solution. Explain your reasoning for your choice for each pair. a) 0.10 M dimethylamine ((CH3)2NH) or 0.10 M ethylamine (CH3CH2NH2) (2 pts) b) 0.10 M carbonate anion (CO32) or 0.10 M cyanide anion (CN) (2 pts)

Answer 1

Here the added HCl is a strong acid and all the given bases are weak bases

Now when w strong acid is added to a weak base (HB), then it form a conjugate acid salt HB+

H+ + HB -----------> HB+

Thus due to addition of strong acid, amount of HB form the solution decreases and that of HB+ increases to the same extent.

Now the pOH of such basic solution is calculated y Henderson-Hasselbalch equation

pOH = pKb + loh [salt]/[base]

And pH = 14 - pOH

pKb for different weak bases are always fixed.

Now the concepts are-

• Higher the pKb, weaker is the base and vice versa.
• Weaker bases have lower pOH, thus higher pH
• Change in pH for weaker bases are more than change in pH for stroner base.
• Or base with higher pKb will have more change in pH than bases with lower pKb

Based on the above concepts, for

1-

(CH3)2-NH = dimethyl amine, pKb = 3.27

CH3-CH2-NH2 = Ethyl amine, pKb = 3.35

Thus upon addition of HCl, change in pH for CH3-CH2-NH2 > change in pH for (CH3)2-NH

2-

CO₃^2- = Carbonate ion, pKb = 0.999

CN- = Cyanide anion, pKb = 4.60

Thus upon addition of HCl, change in pH for CN- > change in pH for CO₃^2-

Lets prove this by taking an example-

Saltte ). 7ar Thm N2 20 ts take saktf ) Ten Bare 3 27 t la 3-27 t (0 6a8) 2.572 4-2.52 ॥ प४ ot tel to tom mre Nk

Herce t amont t solut de crese to t same lo mal ef ndded lets 20t to - 30 (sadH= Thenme Cese to o - lo -10 Norp - 3. 27 tl( 3.27 t (-0 17 2 793 4-2-93= . 207 p cAange np(. 42e 221

Nas n pk ot 2 ph C Te saly sal CBare) Thun ГВan) NHY Then p- pks le salt bane 3-35t = 2.35t(-0 g 23) 2527 = 173 -14-2.C27 amougtof Aher Hel r odded same are t de creane and 3at creare Game lomel ot t dded

The ete salt cflo 2 Csat) 3- 3 s t( o.55) 2.8 (4-2.2 = II-2| PH 0 273 change p foC-CNt change np
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