Question

Calculate each Ka using the information below.

Acid Used: KHP

Molar Mass of Acid: 204.22

Amount of KHP used: 1.04 g

Initial (mL)	0	2	4	6	8	10
Final (mL)	2	4	6	8	10	10
Amount Added (mL)	2	2	2	2	2	-
pH	4.24	4.27	4.30	4.46	4.51	4.63
Ka

Answer 1

Amount KHP Used = 1049 no o moles of KHP = 1.04 mole 204.22 vol. KHP Used for each case - 2 m2 = 0.0022 IKHPI = 1.ou. zou.R2 / 0.002 = 2.54523 M For weat acid [H30+3 = Tkatkap). log + H307] = log ka + = log tkHP). DHE À pka - log T RHPJO log IKHP3. = 169 2:54523 = 0.4059 I Pka = PH + 0.4059 pra= 2 pH + 0.4059 For PH 4.24 pka= 2x424 +0.4059 = 8.9859 Ka= antilog (-0.8959) Kan 1. 3X10-9 For pH = 4.27 Pka= 2x 4.27 to 4059 = 8.9459 Ka= antilog (-o.g459) Ka = 11T3X10-9

for PH= 4: 30 pka= 2 4.30 + 0.4059 = iso 59 ka- antilog (gig059) = 9,855 xlo-lo for PH= 4.45 pkg= 24.46 + 0.4059 = 9:3259 ka - antilog (-9.3259) = 4.72 x 10 for PH= 4:51 pka = 2 x 4:51+ 0.4059 = 9.4259 kas antilog (-9.URGI) = 3.75 X1o-lo For PH = 4.63 Pra = Z x 4.63 +0.4059 = 9.5559 kas antilog (-9,6659) = R: 16x10-10