Question

suppose it is required to make a buffer solution of pH =4 using acidic acid and sodium acetate.How much of sodium acetate is to be added to one litre of N/10 acidic acid? HInt:disassociation constant of acetic acid=1.8*10^-5 M

Answer 1

(Let sodium acetate’s Molarity is M)
Here,                HAc   + H2O  = H3O+  + Ac-
Initial                 0.10                  0          M
Change             -x                     +x        +x
At equilibrium    0.10 – x           + x       M + x                                                                                               
                     [H3O-] [Ac-]              
Now,   Ka  = -----------------                                                                                                    
                      [ HAc]                       
 
                                   (x) (M+x)             
                             = -----------------    =   1.8 X 10-5     
                                  (0.10 – x)
Now, it comes roughly (x)(M)/(0.10) = 1.8 X 10-5;  -(i)

neglecting x in M+x and 0.10-x;

Here  pH = 4 (given), or x = 10-4,
putting the value of x in eq (i),

We get the value of M = 0.018M

So, sodium acetate required = 1.476 gm per litre
Answer:  1.476 gm in one litre

Answer 2

Explanation:

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