106Q1. use buffer equation, as this is acidic buffer pH = pKa + log([salt]/[acid])
pKa = 10 - log(5) = 10 - (1-0.3) = 9.3
log([salt]/[acid]) = log(0.15/1.5) = log(0.1) = -1
So, pH = 8.302
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
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10 Answers
106
106in that case I can only think of this
HCN = H+ + CN-
(1.5/0.5) 0.15/0.5
(1.5 - a)/0.5 a/0.5 (0.15+a)/0.5
Ka = (0.15+a)*a/0.5*(1.5-a) = 5*10-10
find the value of a.
Then find [CN-]/[HCN] and use this new value in the Buffer equation
1357@ asish.....After finding 'a' do we really need buffer equation?? :)
1357concentration=no. of molesvolume of solution in litres
here the volume is total volume only
24Q1 1dm=10cm
=>1dm3=1000cm3
=>vol here=500cm3=500mL
[HCN]=1.5/500 *1000=3
[KCN]=0.15/500 *1000=0.3
pH=pKa +log[Salt]/Acid]
=9.3 +log(0.3/3)=8.3
=> a)
Q2 a (becoz on adding such small maount of acid to a large amount of buffer pH change will be negligible)