106Q53. 1 ml of 0.1 N Hcl is added to 999ml solution of Nacl The pH of the resulting solution is
moles of HCl = NV (n-factor=1)
= 0.1*0.001 = 10-4 moles
volume = 1L
so [H+] = 10-4 moles/lit
Hence pH = 4
96) 50 ML of a solution containing 10-3 mole of Ag+ is mixed with 50 ml of a 0.1M Hcl solution. How much Ag+ remains in the solution? Ksp of Agcl = 10-10
a)2.5 * 10-9
b)2.5 * 10-7
c)2.5 * 10-8
d)2.5 * 10-10
59) Wats the ionization constant of an acid if the hydronium ion concentration of a 0.4 M solution is 1.4 * 10-4M ?
79) 0.1 M formic acid solution is titrared against 0.1 M NaOH solution. What would be the difference in pH b/w 1/5 and 4/5 stages of neutralization of acid ?
a)2 log 3/4
b)2 log 1/5
c)log1/3
d)2 log 4
84) The pH of the resulting solution of 20 ml of 0.1M H3PO4 and 20 ml of 0.1M Na3PO4 is
a) pKa1 + log2
b) pKa1
c) pKa2
d) (pKa1 + pKa2) / 2
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8 Answers
106
135796)
Ag++Cl-→AgCl
mmol 1 5
0 4
AgCl→Ag++Cl-
0 4
at eq x 4+x
(x/100)*(4+x)/100=10-10
or x=2.5 x 10-7 millimoles
135779)
pH=pKa+log([salt]/[base])
pH1=pKa+log(0.02/0.08)
pH2=pKa+log(0.08/0.02)
find pH2-pH1
1357btw why you deleted Q 53 from your questions
You should not do that as it is always answered.
So others can get some help from your questions and understand what asish has written for
1357Q 96)
First I assume that the reaction will go to completion(Ag+ and Cl- will react to form AgCl)
so mmol of Ag+ and Cl- will be 0 and 4 respectively.
Now AgCl will dissociate to Ag+ and Cl-
and so the result follows
