Question

LUIUS. Question Completion Status: QUESTION 21 A 35.0 mL sample of 0.180 M benzoic acid is titrated with a 0.180 M NaOH solution. What is the pH at the equivalence point? The Ka of benzoic acid is 6.3 10-5 11.20 9.80 4.20 08.58 7.00 QUESTION 22 A 25.0-ml sample of 0.150 M hydrocyanic acid (HCN) is titrated with a 0.150 M NaOH solution. What is the pH after 13 ml NaOH has been added? The kg of hydrocyanic acid is 4.9 x 10-10 9.35 ООО 8.75 8.49 9.28 ОО 3.15 Click Save and Submit to save and submit. Click Save All Answers to save all answers. Sa
i need help on both

Answer 1

Answer. 21 benzoic Aid = Каз ska for 6.3 хуб Рka = -1 ба ko pkas - Јоа (зхо - 5) Рka = — Iva 3 - Iva 1° rko 2 - 0 799 + lоа го Ч. 2 At the stoichiometric point, Cacid] =[base [acid] =) [base] Using Henderson- Hasselbatch Equ_ PH = ko (base] log Cacid] dog! РИ - 4.2 4. 2 - 0 PH (PH = pH = 4.2 Answer. 228 ka NOW, = 4.9 хр! fka - — loako pka =- До 2 ( 4-9 х 16 10) Рka = - Loe 4:9 + 10 Рka -- 69 + 10

pka = Maced nbase [Acid] - [ Base] Mbax/v NaoH corresponds 13mL addition of addition of L Molarity of Х 13 x 10 Naon non- 0.15 males IL 13x10 MOH- х -3 moles 1.95 x 10 Moh- of HCN.- 3 x Initial amount malarity of 25 X 10 L NHCN HCN. L Х 0.15 moles IL 25 X10 NHCN -3 moles. 3.75 X 10 Мис N after the so, the remaining amount Naon of 13 mL -3 moles امامه 1.9Jx10 3 -3 x 10 (3.75 NHCN moles -3 1.80 x 10 MMCN = -3 1.80 x 10 motes MHCN nauid non- nbase 0.92 ngeles

Eqh Using Henderson Hasselbalch PH pka СВО log [ Acid n And РИ 93 Тоа л base РИ 9.3 Loa on 2 РИ 9.3 (— О. 0362) ss (на 9.3