Question

Page 1 of 6 a. (3 Points) If 145 grams of lead nitrate were added to water to make 1.50 x 10 mL of solution, what would be the molarity of the resulting solution Pb(NO) molar mass w 331.21 gmol b. (3 Points) What is the molar concentration of chloride sons in a solution prepared by mixing 50.0 mL of 1.5 M KCI with 100.0 mL of a 2.5 M Cacla solution? took 100% Point) A 147 mil portion of an HCH Solution was titeated with BSM NOH ml of the base to neutralize the sample What is the molarity of the HCI

help!!! how do I answer this

Answer 1

Melecular weight of Pb(NO3)2= 33). 21 g/mol. Weight of leac nitrate used= 14.5 g Amount of water used → 1.50x103 ml. 1 103 ml = I litre Molarity of solution = weight of solute Molar weight x Volume of solute of solution in litre. » 145g 331.21,9/mel * 1.50x103 me → 145g x mol 331,219X1.50 litre 0.292 mol litre Molarity of lead Nitrate solution → 0.292m bi Volume of KCl solution given = some Concentration of kee solution = 1.5m

Volume of Calle solution given = 100.ome concentration of Cacl 2.5m No. of moles a molarity a ralume:, Ree - kt + ceo روف روی No. of moles of Kce = 1.5 mx 50.0 ml 1.5 mol x 50.0 ml. litre = 1.5 mol %50.0 ml = 0.075 moles, 1000.0 ml (in moles) Call - law (in moles) cart + acco CGQ ccm

No. of males of Call, → 2. SMX 100.ome - 2.5 mol x 100.ome litre 25 mal x 100.0 0.25 moles 1000 ml from inic equation, imole of Calle giving two moles of ceo ions. No. of moles of ceoins = 2x No. of. moles of Calla = 270.25 = 0.5 moles, Total No. of moles of ceo ins 0.5 males + 0.075 moles => 0.575 moles. Total volume of balution = 100.0 me + Soome . 150.0ml.

Now, malasity = NEO - of solute Polume of solution in litre 0.575 x 1000 llooomb = . 150.ome I litre) 3.10:575 X 1000 = 3.83m 150.0 me the Molar concentration of ceo ins= 3.83M © Volume of He = 147 me = vi Molasity of NaOH given = 0.625m=Mq Volume of Nooh used = 100.0 ml = Vz No. of moles of Hae ubed = ? No. of males of NaOH used - 0.625m x 100.0 ml 90.625 moles x 100.0 ml litre 5 0.62 Smely loo ml = 0.0625 moles To com

Now: Hellad + NaOH(aq) Nall & H₂O 1 . I mole of NaOH neutralizes Imole of Hl. So. I mole Hel = 1 mole NaOH. By molurity equation, let my be the molarity of He. in M, avi = M2 X V2 M, x 147 ml = 0.625mx loome M, 7 0.625 x 100 ml. M. 147.0 ml My = 0.425M Molarity of Hae Dabution = 0.425 m

d. Volume of HCl 150 ml = 1 Molarity of HCl. → 0.250M. = M, No. of moles of HQ = M, XV, - 150 ml x 0.250M P 150 ml x 0.250 mol S titre 150 ml x 0.250 mol Tat t oou me 0.0375 moles Volume of NaOH = 70 ml = 1/2 Molarity of NaOH = 0.375 M = M2 No. of mules = Max Ve 3 1000 ml 70 x 0.375 moles 0.02625.males 000 0.02625 moles com meutralise only 0.02625 moles of Ha. Remaining HQ 10.0375 -0.02625) ► 0:0!125 miles,

Total volume of solution now:- 150 ml + 70 ml a 220ml. Molarity = 0.01125 möb-x 1000 220ml = 0.01125X1060 M = 220 0.051m Mularity of remaining solution = 0,05174 ( Plz like)