Question

pH, INDICATORS AND TITRATION CURVES PRE-LAB ASSIGNMENT: 1. A certain indicator is blue at pH = 4.0, yellow at pH 6.0, and red at pH 8.0. a. How many pKin values will this indicator have? (Assume all its colour changes are given.) b. A solution with pH = 5.0 has some of this indicator added to it. What colour is the solution likely to become? c. What are the approximate pKin values for the indicator? 2. A 0.20 M solution of a certain weak acid has a pH of 4.50. What is the Ka of that weak acid?

I'm having trouble with 1a, 1b, 1c and 2...

Answer 1

II The inoueatar has, 3 Colourat 3 different PH. That mease the indicatar has 3 täfferent farms at the 3 PH.. a) Suppone, Heln - A pH = 4.0 – Blue Hin . At pH=600 - Yellow In2 - At PH=80 → Red Hence Hain HintHt Acearting to the Hendernan-Hasselbalch equatian, pH = PITA+lay L Hint [H2 In] The eclawr will preferably change when Chin is 10 times to the cancentration at the [Hz In] ar viceversas Hence, blue calaur when i [ H2 In] = 10 [HIV] - TPH = pleme-11..) Yellow colawr when, LH -] = 10 ( H In] TPH PK + 1 .1 Similarly for Hin - In2r+H+ Accarting to the Hanelbalch equatian, PH = p<th> tiny [ In2- Chini The yellow calaur will appear when, Hin] =10€ Iner] af PH-Piha -1 .. The red colour will appear when (In 2] & 10[ Han ] al pH = Pkm2 +1 ...) Since there are two equilibriums at the intrucatar there will be two kan values and hence of wo pkin value

At PH=5.0 pH = Pkinit [H₂I] when, pH=4 That means when, pH= 5 4=pkinin that is in between PH=4 and pH=61 when PHEG Canelusian . pH = pkin, (= pkint! That means Hence, the two different. lag [ Il In] =0 Parm with two different colour will be at lilrario. [Hain The yellow campauend and (Hin] blue campaund are at lil [H₂in ] ratio an [Hin] = [lla In Hence of he nolutian cuill have greenish colour c) The indieatar have ofwo p kan value, Pking and PR Inza P = = [ PH, + H ] - [ H + 1 = 5 Pkine= [ PH₂ + PH3] [6+8.] = 7 2 Far a weak acid HA HA Y Htta" c(1-2) ca ex (ca) (ca) elend) ka = [H+][A-] [HA] ka= @²x² as a <<< ka = caz New CH +] = exs.3.16 x 10-5 M = 0.000158 Naw Ka=ex? Ika=5810-9