Question

8. 25.00mL of 0.100M of a weak acid HA is titrated with 0.100M sodium hydroxide:
Ka = 1.8 x 10-5
Calculate the pH of the solution after the following:
a) Prior to titration (no NaOH added yet)
b) 7.50mL of NaOH added
c) 12.50mL of NaOH added
d) 25.00mL of NaOH added
e) 32.50mL of NaOH added

Answer 1

a RH 10 omi Naon & HA = ut ta o loomo o ox x 0.100* t neglect x as Ka is clos Kaa CHT) (A) CHA) 1.88105 x² o-loo 2) x2 Cort) a 1-3x103 m. pH log Cut) [2.89]. J.soml woont mores naon added & 0.100m x 7-50 ml x 103 2 0.750 x 103 moles. 2 o 750 mmoles moles I HA 2 0 100m x 25-oome x 1031 = 2.50 x 103 moles 2 2.50 mmoles HA realts Noon , form A (consugate base) 6) with тон HA & Naon - Ä Nat & Ho 2 2.50 mmoles 0-750 mmoles o C -0.25 -0.75 to-750 1-75 mmoles 0.750 mmoles. AS sownon contains weak acld and its Sourlon Conjugare at acts as butter pka z tog Ka 2 10g (1-8x100 24.24. pt a plea t 10g (A) 2 To frommeles 4.74 + 10g) 4-37] 5 mroles :

6 12-50ML Noon added noon moles added 2 2 z 0-100m x 12.50ml 1.25 mmoles moles A formed moles HALAL 2 2.50 -1.28 = 1-25 mmoles moles Ha è moles à formed. Here so it is half- equivalence pka point (log (A) a logl zo) pH 214-74 25-00 ml NaOHL moles Naon added I 0-2.50 mmoles z moles HA cinitia) point, all ha converted so it is equivalence A moles HAONH) - a to Toto von 0-100m x 25.00ml 25-00ml + 2500m EA) = 0.0500 m. Calculate Ko as e acts as base Kor KW 2 1.0x10th a 5.6 x 1610. - T-8x 107 A + H₂O AHTOH т оогоо G 0-0500 negled & as kiverless ky 2 CAD Corr) - Cont) x - J kg (A) 255-6x100x 065oom 2 Cort] 5-3x100m - pon - 10g (5-3x156) 215-28] pz 14-00 - 5-28 2.18-72) 22-50ml Nasty other equivalence point, So noon excess. Cortlieft & Toral moles - moles react 0-100m (32.50 sochel Total volume (32.50 +2500) 20.0130 m.

PM -10 C0-0130) 2 1-886 1 2 14-000-1-886 2 2-114