Question

1. Hydrogen cyanide (HCN), sometimes called prussic acid, is a colorless, extremely poisonous and flammable liquid that boils slightly above room temperature (25.6 °C). HCN is produced on an industrial scale and is a highly valuable precursor to many chemical compounds ranging from polymers to pharmaceuticals Determine the pH of a buffer solution is prepared by mixing 125 ml of 0.080 M KCN with 200 ml of 0.025 M HCN. The Ka of HCN is 6.2 x 10^-10

2. Determine the pH of a solution prepared by diluting 50.0 ml of bleach, a 4% w/V solution of NaClO (sodium hypochlorite), to a new volume of 250 ml, if the Ka of HClO is 3 x 10^-8.

16. Exactly 25.00 ml of 0.0685 M HClO₄ solution is placed in a flask and titrated with standardized 0.050M NaOH.

        A. Determine the endpoint of the titration

        B. Determine the pH of the solution after the following amounts of NaOH are added to the acid solution:

                       a.) Initially, 0.00 ml NaOH.                   b.) 5.00 ml                                c.) 15.00 ml

                       d.) 25.00 ml                                              e.) 34.00 ml                             f.) At the endpoint of the titration      

                      g.) 34.35 ml                                              h.) 45.00 ml                             i.) 50.00 ml

    

Answer 1

10 Wore product When 125 mL of 0.025 (M) 0.08 (M) KON is HON; the 1) KCN is mixed with 200 ml mi be 3 CM) HEN; the considered. Final dilution factor shall volume 125+ 200 B = 325 mL If Shen be the strength of Hcn in the Thus ; if shen be the strength samble;. SHONX 325 = 2008.0.025 or; S Hon : 0.0154 (M) if SkoN be strength of diluted SKEN X 925 = 125 x 0.08 or; Sken. 0.0308 (M) Thus; from Hendersson equation, pH = pkat log [salt] Here salt=ken Tacid] and how on, pH = Rk - log 10 16.281010) & log 0.0308 0.0154 or. pH = 9.207 + log 2 = 9.50%

0426 moles on 100 ml, Nood present : 49 on 1000 mL, Naol present = 40g - 40 moles mwaoce 0.537 moles Thus; 50.0 ml of 0.537 moles is diluted to 250 ml. Hence, its new strength is 0.537 moles - 0.107 salt is known. RD Neons, Thus concentration of neai denboraron oef Namente para + 200 HU NaOH + Hou Naod & H2O H20 - Hoot tu IM) Naod shall give Thus 0.107 0.107 (M) Holl d Now : Holl - 0-107 11-0) H t + cor 0.1074 0.107a degree of dissociation 22017 0107 Ke 10.109 2 ) - 0.107d? [ 0.109 (14) orka 0.10742 or., da v ka [ H+] = 0.1072 av Ka.xo.107 [H+9= 3.21610-4 pH = 8.49

6. (A) det NaOll be the volume required neutralisation Nou X 0.060 0CD. 25 x 0.0685 or [unnoll 3 u. 25 mL a endpoint 1B) (a) when no NaoH is added ; [+] = 0.0685(M) 5.00 added to 25 ml, ml NaOH il when (6) when 5.00 mi na o total volume = 30.0mL Helo 25X0.0685 m - 0.059 (M) NAOH = 5:00 X 0.050 - 0.008 (M) .CA 0.0496 S Halo PH : 1:3) 1 volume : 40.0 mL when 15.0 mL added 25% 0.0685 (M) = 0.00 28 (M) 40 54 0.050 (M) = 0.00625(M). S Naolt Naolt = - no Soon = 0.03655 pH = linu Id) when 25.00mL added, half-dilution shall occur. SHclou: 0.034251M) I saw Snach = 0.025 (M) Cuore PH: 2.03 но собоб

( When 3 uoo mi added volume : 54 m2 Huldu Snadly 0.029 (M) 0.021 (M) Shelono 0.008(M) (1) At end-boint, volume 59.25 mL HUO 0.006(M) " SHuvu 0.0290 (M) I SNAOH. 00211(M) PH: 210 59.35 m2 (9) when 3 u. 35 mL added volume = PH = 2.10 [ calculate as before ? th) when u5.00 mL added, volume = 70.00.ml SHulon = 0.024 (M) 2.. .: Inaon . 0008(M). s Naoth = 0.032(M) ] Here all acid shall be consumed. Thus ; pot: 2.09. PH: 11.91 [ " PH + poH : 147 (d) for 50.0 mL Shilom. addition; volume = 75.00 m2 X 0.0699 (M): 0.0229(M Snaoit : 2 x 0.050 (M) = 0.0333(M) SNOOH = 0.0100(M) POH : . 1.98 :|PH = 12.02