Question

PT - 1L 5. K, for hypochlorous acid, HCIO, is 3.0x10-8 Calculate the pH after 10.0. 20.0, 30.0, and 40.0 mL of 0.100 M NaOH have been added to 40.0 mL of 0.100 MHCIO.

Answer 1

ANSWER: NO Ka = 3.0 x 10-8 pka = – tog ka = – log 3x10 OR pka = 7.52 millimotes of HUO = Molarity * volume = 0.1 X 40 = 4 mmol • When 10 mL of o.l M NaOH is added mmol of Naoh = 0.1x10 = 1 mmol Acid is partially neutralised. Finally, mmol of acid left = 3 mmol 4 mmol of salt formed = 1 mmol This becomes 'acidic buffer'. Using Henderson - Hesselbalch equation pH = pka + log [salt] lacid] = 7.52 + log = 7.04 3 But since it is acidic buffer, pH cannot be above 7; we need to consider t ions furnished by water. Therefore

ka = [H"] [ Anion) 3x108 M']*! [ id] [w] 9 x108 Actual [ut] = 107 + 9x108 (107 m from water) - 1.4 x 107 M pH = -log 1.9x107 = 6.72 is added When 20 ml of o.lm NaOH satt formed = 2 mmol. acid left = 2 mmol so Ka= [H +] x 2 = 3x108 2 Actual [at] = 107 + 3x10 8 = 113610-7m pH = -log 1:3x107 = 6.89 - when 30 ml of Naoh is added salt formed = 3 mmol acid left = 1 mmol so Ka = [ht] x 8 = 8x10 8 [ht] = 10-8 M. Actual [nt] = 107 + 10% = 1.1x107 m

pH = -log 1.1 x 10 = 6.96 when 40 mL of Naon is added, entire and is neutralised & only salt remains. we use the concept of hydrolysis of salt of weak acid & strong base : [ht] = Kwka у С 4 product of water) Here & Kw = 10-14 c = cone (ionic of salt = u mmol 80 mL = 0.05M » [at ] =/10 14 x 3x10-8 = 7.74 XIÓ" M 0:05 so pH = -log 7.74 x 10 = 10.1