Question

(a) Based on the measured [HC4H4O6−] and the molar mass of KHC4H4O6 (188.18 g/mol), what is the gram solubility (g/L)of KHC4H4O6 in solution #2 trial 1? Answer to 2 decimal places.

Assigned saturated solution (circle) Solution Concentration of NaOH solution: 0.0259_M Concentration of KCl in Solution 1 0.0150_M Saturated Solution | Saturated Solution II KHC4H406 KHC4H406 with added KCI Trial 1 Trial 2 Trial 3 Trial 1 Trial 2 Trial 3 25.00 25.00 25.00 Volume sample solution 25.00 25.00 25.00 (mL) 0.00 0.00 Initial buret reading 0.00 15.00 (mL) 32.86 32.91 Final buret reading 24.82 39.97 (mL) 32.86 32.91 Volume NaOH 24.82 24.97 (mL)

(b) TRUE/ FALSE: The Ksp of solution #1 and solution #2 should be the same even with the add KCl.

(c) TRUE/ FALSE: Using LeChatlier’s Principle, the molar solubility of KHC₄H₄O₆ should be the same for solution #1 and #2.

Answer 1

The insoluble satt for this experiment is potasimsium hydrogen tartanate , * # Cy Hy 06 The solubility equilibrium of K4Cute Ofis KHCC H4O6 (5) & Krlag) + HC 4 H 405 69). kap = [*] [ HC41406) -- In absence of any other source of potassiumion the CK+) will equal to [HC 44406%. So determining the [Heute 06-7 by bare Nao A acill allow to determine koß. oshe reaction aciti Noot or of & HCyphy Of ": Hey Hy 06 (ar) tot (ar) y 4406 Tar) +4₂0 (4) - lü) e presence of potassiumion by adding kel (soluble) the value of kop will be kap=t&I tot [HC y H4O6 ] In Both cares Kop expression is same Roller [HCyte, 0602 I no added kcl [Ktot = [Hlytulo) determined by de termined I [kator = [Acy Hn067 with addedke + [koll by Noot Saturateso/ Naolt

of 0.0259an to titrate For saturated solution I Trial 1 volume of noot used 32: 86 Naolt Number of endles of ott used Heyty of = 0.0259 Mostert & 1 and worst t x 1000 24 x 32 86 - 8.51810-4 mol Olt - Now, according to stoichiometric ratio of the reaction (1) mot ot-react with Imol HCyty Of .8°SIXIOO u 8:51X/6 now, volume of saturated spotion of KHC, H40s was taken 250m . [Acthy ord = 8.51810no 103 25.0m - 003409 here, Ikht= [HCyty Os] = 0.03 404 kob = [HC, 0405] x [ kiter = 003404X003904 = // 6X/03 Triale similarly Trials mol of oth recent authAlytteoon = 0.0259 m x 320 1 ay = 8.52x10 mol 8.52 x 10 I 4cyty 067 = 25 = 0.03408 M : Qt, 0.0940300 - Kap = 0.09408 X009408 = 116 x - 3

(a) der solution #2 trial gram solubility of Kle, Hy86 is. 4.8ngh (e) the kop of solution #1 solution #2 show & be same eron cuith add Kol = Fake (c) using Lechatelier's principle, the modr sulebility of KHC y Hushowd be same for solution #24 #2 - Fake I Because due to comman con effect of tt in dotion 2 the solubili & mill decreased