Question

PART A. Determine the pH of a 9.553 mM weak acid solution that has a pKa of 8.83

PART B. Calculate the pH of a 319 mM weak base solution with a pKb of 9.04.

PART C. Determine the pKa of a weak acid solution that has an initial concentration of 0.334 M and a pH of 4.54. Hint: You know the pH of the solution, so you can easily determine x in your ICE table

PART D. Calculate the initial concentration of a weak acid (pKa = 7.53) that is needed to create a solution with a pH = 4.54. Report your answer in mM.

PART E. How many moles of Mg(OH)₂  is needed to neutralize 66.71 moles of HCl?

PART F.What is the resulting pH when 0.015 moles of HCl is mixed with 0.0272 moles of NaOH in 0.48 L?

PART G. What is the resulting pH when 14.75 mL of 0.59 M HNO₃ is mixed with 14.83 mL of 0.14 M NaOH?

**PLEASE ANSWER ALL PARTS**

Answer 1

А. Foi weale aude, [ist] Kac -pka а - 10 8.83 То -4 1,5х10 caq.6 33 мм 9.553 xio Mi [*]: ) -5xir9.553x ° 16.9.553жіб® : ) 4-33xts'- 3-74x5 6 үг -log_н+7- lost3-74x156 : - los (3-79) +(65log 107 г. [о. 58 -6] = -(-5-kj =5, g. -3 C: 319410 Lov-2: JEьс Icb= 10 9. oh9.2 х 1° for: Jq.mxm+захіб? - 246 У по —РК, 9. op = to -2. 11 №10 ро» -Los for-1. -Log (1x156) - [tosti+ (-62L010) - [). 23-67 г--4.25 = 4. 8 pH = )-рой - 14-4. &7 9.2.

с. се о, 2 , н: 4.54 - 54 jo = 2,88 x 10 үн. [ht] = 1 [+]: [ка с <a Сб. 339) эка: &•3х10 -fo О, 33y (2.&& x69) -10 7 24, 85х10 = 2.49 хіб 4 x 9 -pka - 7.52 ka=10 = To р. 2,45 хто x108 -4.С 5 = 2,88 x jo = 10 Lu*], -н. L++7 - кас D+7”. Кас Le+j”. (2.88x155 2.45 куб 2D CE ka 8.3x 0 2.40xo 2.81 х2 = 28-1 хон : 28:1 мм MOcl, +20 Е. 2 Hel + Mg(OH) moles of Mg104) = 2 x moles of Hel 2 x 66.1 2 133.42 оlеѕ оf w/

.م Hol + NOOH nad tho 그 0.015 0.0272 to.015 -0.015 -0.015 O'015 0.0122 moles of Naoh remaining = 0.0122 mol [0] = molese of OH = 0.0254 volume of Station (0) 0.48 0.0122- Рон Hog Coways -log(0.0254) =-(-1.62 = 116 12.4 PH = 14-POH = 14-1.6 = a. Naoh t Nanoz it to HNO3 0.14x14.83 1000 = 0.0021 noles of NaOH= Molarity x volone (ml) 1000 moles of HNO3 = 0.59*14.75 0.0087 1000 NONOZt to Naot & HNO3 I 0.002) 0.0087 - to.0021 -0.0021 c -0.0021 0.002) F 0:00 66 0.0066 trovo moles of Htcom kwes remaining = 0.0066 [++] = molete x1000 onab Totalvolomelo) (14.715+14.8V 0.0066x1000 0:22M 11 29.58 PH - rog Cit+] = -loglo.29) =-(-0.66) = 0.66