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Q. 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

Equilibrium

Solution:

Buffer (A):
$ pH _1= p K_a+\log \left(\frac{x}{y}\right) \quad \ldots \text { (i) }$
Buffer (B):
$ pH _2= p K_a+\log \left(\frac{x}{y}\right) \quad \ldots \text { (ii) }$
Since $x>y $
$\therefore pH _1- pH _2=1=\log \frac{x}{y}-\log \frac{y}{x}$
$\therefore 1=2 \log \frac{x}{y}$
$\log \frac{x}{y}=\frac{1}{2}=0.5$
$\frac{x}{y}=\text { Antilog }(0.5)=3.17$