Question

2.         The Ka of nitrous acid HNO₂is 4.5 x 10^-4.  In a titration, 50.0 mL of 1.00 M HNO₂is tit  rated with 0.750 M NaOH.     

            a. Calculate the pH of the solution:

                        (i) Before the titration begins.

(ii) When sufficient NaOH has been added to neutralize half the nitrous acid                   originally present.

                        (iii) At the equivalence point.

(iv)When 0.05 mL NaOH less than that required to reach the equivalence point  has been added.

(v) When 0.05 mL NaOH more than that required to reach the equivalence point has been added.

            (b) Can Bromthymol blue be used as the indictor for this titration? Explain your answer.

            (c) Will methyl red be a satisfactory indicator here?

3. A 0.500L solution contains 0.025 mol Ag⁺.

(a) Calculate the minimum mass of NaCl that must be added to precipitate AgCl from the solution.

(b) If excess Cl^-is added to the solution the AgCl precipitate dessolves due to the formation of  [Ag(Cl)₂]^-= 1.8 x 10 ⁵.  Calculate the minimum amount Cl^-that must be added to dissolve the precipitate.

-we are given the weak acid and the conjugate base in the problem. we need to go from there. First find the pKa for the acid then proceed from t

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For number 2

Start by writing the balanced chemical equation.

then proceed as we did in the lecture slides to solve for pH.

For number three

this is a Ksp problem

We did one of these on the lecture slides as well

Again start with a balanced equation then write a Ksp expression for the reaction.

yu can then solve for [Cl-]

Answer 1

solution :- Ka of Nitrous Acid = 4.5810-4 ANO2 + Naon Na NO₂ + H2O moles of Naona males of HNO2 MINI = M2 V2 0.750m xvi = 1.00M X 50ml. VI 50 ml. 0.75 VI = 6.6.67 mei Volume of Naon at equivalence point = 66.67 mL. رن ph of HNO2 solution: Holor is a weak acid 어 가 + NO2 INO2 0 1 1.0 k de -x C. x re 1.on ka 14+] [N102] (MNO2] 4.5xión y W 4.58104 4.5 810 -u? * +4.5X1059x. -4.5.x104 = 0

a= -b I ſ b²-4ac 2х4 k -Ч: 5 xo 9 4 2-625 xot-4xi x+(-4.5 xis") 2. ។ x = -4.5 X10 + 2 - 025 Хо +o-ool" 2. ч. 5 Хо Ч+ -Оч?ч 2. thiy posihve sigh + о. Ч292 4, X ч: 5 xo-Ч . ооч1412 = 0:02090 2. - Х- | ht) 0:02 098 м. ,, ри - - alu"] = - Pa Co-42954) ри 1-(14 1 - ): 64 (i) At half equivalence point ph = pka РИ - - ") ka P1= - 1а (4-5 хв) [P = 3-35

there will be only At equivalence point NaNO2 in the solution, maley of Nano2 formed = moles of HNO2 50 mmoli Total volume of solution = Бо hl + (- 67 , г) 6. 6 ml. NaNO2) mol Volime 5о мало. 16. 62 L. О-Ч2 +6° мо0 JL + pha 2+ ) РЕ, Na NoL is a salt of weak acid and strong base ph of solution is given by I loge where ca ти?vity % + + + x 3-35 + х = (o4284) ри с 1 + 1-475 +2 х (-о: 3679) ри = 9-615 - о. 189 ри - Salta ри — 2 3:49

SIN 0.05 ml Naon less that that equivalence point; 66.62 ml of Naon is added; V= 66.67 -0.05 = 50 nmol. initial male of NO 2 = moles of Naok added - 0.750 mx 66.62mL 49.965 mmal HNO2. = 50-49.965 mole of remaining 0.035 mmal. male of Nash hale of Na No2 formed 49.965 mmal. pn = pka + log [NaNor) [ Hana ph = 3.35 + log (49.965 0-035 ри 3.35 + 3.15 ph = 6.50 0.05 me extra Naon added after equivalence poinh V = 66.67 + 0.05 = 66.72 ml. call HNOL will be heupelized, excess male of Nadl= 0.750 m X 0.05 mL = 0.0375minal. total volume = 50+ 66.72=116.72 me.

[ Naon] = 0-0375 mmal. 116.72 me. 0.000321 malah 0.000321 mal/ (ou) (44) 1.0 x10.14 0.000321 (H+) (n+) 3.112 x 10" = 3112.53X10 14 M log (ht) leg ( 3.112 X10 10'' 10:51 ph= PH 7:1 b blece = transition range Bonthymol blue, PKI puatequivalence point = 8.49 for Brom thymal pki z ie 6.1 to 8.1 Bromthymal blere can not be used as indicator in this titation: Methyl jed : pke = 4.95 fransition range 3.95 to 5.95 Methylted also can hat be used;