Ionic EquilibriumMedium
Question
Two buffer solutions, A and B, each made with acetic acid and sodium acetate differ in their pH by one unit, A has salt : acid = x : y, B has salt : acid = y : x. If x > y, then the value of x : y is
Options
A.10,000
B.3.17
C.6.61
D.2.10
Solution
Buffer (A):
pH1 = pKa + log(x/y) .... (i)
Buffer (B):
pH2 = pKa + log (y/x) ..... (ii)
Since x > y.
∴ pH1 - pH2 = 1 = log (x/y) - log (y/x)
∴ 1 = 2 log
log
= 
= 0.5
= Antilog (0.5) = 3.17.
pH1 = pKa + log(x/y) .... (i)
Buffer (B):
pH2 = pKa + log (y/x) ..... (ii)
Since x > y.
∴ pH1 - pH2 = 1 = log (x/y) - log (y/x)
∴ 1 = 2 log
log
= = 0.5 = Antilog (0.5) = 3.17.Create a free account to view solution
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