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1 M solution of  CH3COOH  should be diluted to _________times so that  pH is doubled. (Given  Ka (CH3COOH) = 1.8×105  )
A) 4 times
B) 5.55 ×104  Times
C) 5.55 ×106 Times
D)  102 Times


Answer
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Hint: The dissociation constant for the weak acid (like acetic acid) is related to the degree of dissociation and concentration of the solution. However, the relation can be further modified to calculate the hydrogen ion concentration of the aqueous solution of the weak acid whose dissociation constant is known. The relation is as shown below,
 [H+] = cKa 
Where Ka dissociation constant, c is the concentration, and [H+] is hydrogen ion concentration.

Complete step by step solution:
Acetic acid is a weak acid. Its dissociation constant  Ka  is determined by the equation which related the dissociation constant  Ka with the concentration of acetic acid and the degree of dissociation. the equation is as follows,
 Ka = cα2 OR α = Ka c  (1)
The above equation can be rewritten in terms of the hydrogen ion concentration. Such that hydrogen ion concentration in the solution is equal to the  cα  . Thus the relation is stated as follows,
 [H+] = cα = cKa c = cKa  (2)
Take a negative logarithm of the equation (2) .we have,
 log[H+] = log cKa = 12[log c +( log Ka) ] 
Since we know that,  log[H+] = pH  and  log Ka = pKa , we have
 pH = 12[pKalog c ]  (3)
Let's substitute the values in equation (3). We get,
 pH = 12[pKalog 1 ] = pKa2  (4)
We are interested in the value of the concentration when the value of  pH  doubled. We already know the value of  pH in terms of  pKa  . Thus to get the twice of  pH multiply the equation (4) by 2.the value of new pH   is,
 pH  =2×pH = 2×pKa2= pKa
Now substitute these values in equation (3), so that we can determine the concentration  C   which is a concentration of acetic acid required for the double value of  pH  the solution. We have,
 pH  = pKa12[pKalog C  ] 
On further simplifying we have,
 2pKapKa[log C  ] pKa = log C  but, pKa = logKa logKa = log C   Ka = C  
We know that dissociation constant for the acetic acid is  Ka = 1.8 ×105  , therefore dilution ‘x’ times would be equal to,
 dilution  x  = 1C = 11.8×105= 5.55 ×104 times 
Thus to double the  pH  1 M solution of acetic acid we need to dilute it by the 5.55 ×104 times.

Hence, (B) is the correct option.

Note: The negative logarithm taking a step is essential in the determination of  pH as direct formula cannot be employed to determine the  pH . This will direct us to the hydrogen concentration. The hydroxyl ion concentration of the weak base can also be written in terms of dissociation constant f as follows, [OH] = cKb . where Kb is the dissociation constant of a weak base such as ammonium hydroxide.