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Tushar Watts Grade: 12
        
Ques1) The dissociation const of HCN is 5 x 10 -10 . The pH of the soln prepared by mixing 1.5 moles of HCN and 0.15 moles of KCN in water and making up the total volume to 0.5 dm3 is
(a) 8.302      (b) 7.303        (c) 10.302        (d)9.302

Ques2) 10 ml of an aqeous soln of some strong acid having pH =2 is mixed with 990 ml of the buffer soln with pH = 4. The pH of the resulting soln is
(a) 4         (b) 4.10        (c) 3.8              (d)4.25
7 years ago

Answers : (1)

askiitian.expert- chandra sekhar
10 Points
										

Hi Tushar,


1) 1/2 lit solution


    [KCN]=0.15*2=0.3


    initial concentration of [HCN]=1.5*2=3


    therefore


    (0.3+x)*x/3=5*10-10       3-x≈3


    0.3+x≈0.3


    x=5*10-9


    pH=log 5*10-9


        =9-0.698


        =8.302


2) as the acid is added the pH decreases the only option is 3.8


   therefore ans is 3.8

7 years ago
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