Question

5

Qc 5. What is the pH of the solution that results from adding 25 mL of 0.33M HCI to 25 mL of 0.58 M NH,? (K for NHs 1.8 x 10) a) 3.87 b) 4.62 c) 8.99 d) 9.13 e) 9.38 Answer: d 6. What is the pH of the solution that results from adding 15 mL of 0.50 M NaOH to 25 mL of 0.50 M HF? (K,or HF = 7.2 x 10) b) 3.49 a) 3.32 c) 4.61 d) 7.53 e) 10.86 Answer: a 174 P 28 Aa & $ % 7 6 4 3 O P U I Y E T R K H G D FP ব

Answer 1

Amaunt оf HCо a dded Molavily volume 6 33 mmo 2S m 8 2s mmo Amount of NH2 added 25 o58 mmof 14.5 mmot NHA + Inihal 145 mm 2-25 mmot 8 25 mm Change .25 mst -8-25 mmd 8- 25 mm Equi librium G 25 wm 25+25) To tal Volume 6.25 6 2S mmt C SD M. g 25 8 25 mmt m L Given kb b 5- 1-8 44 9.26 14-p 14-44

Henderson- Hasselbalch equal cn Using Salt Cbase + Csalt base H [salt ro Cbase) Cbase) salt + 9.26 toy 6. 25 8 25 H 6-25 9. 26-6.2 y 9-25 lo 1.14 H 7-14 Amswer d

vol me Molavity 6) Amount of Nao H 7.5 mmot Awusunt H F 2 S w X o. 5 mmof htt I 2.S mmo O H +H20 Ini bal 7.5 mmst 12.S mmot 7Smmt 75 mmd - 1 .5 mm Chanfe Eauili brium 5.0 Mm 7.S mm TO te Smm C 4 F) Voluma (51t 40 mL 40 mL C) 7 M. 40 40 mL Given ka 7 2 3- I 4l7 2 :

Henderson- Hassel balch equat Csalr? Lacid P H lon 3.14 H 3 1 P- 3.216 3 3 2 -5 3.14 11 3.32 Answer a P