Answer: The pH of the buffer is 5.25
Explanation:
Let the volume of buffer solution be V
We know that:
[tex]\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution}}[/tex]
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[\text{conjugate base}]}{[acid]})[/tex]
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of weak acid = 4.90
[tex][\text{conjugate base}]=\frac{2.25}{V}[/tex]
[tex][acid]=\frac{1.00}{V}[/tex]
pH = ?
Putting values in above equation, we get:
[tex]pH=4.90+\log(\frac{2.25/V}{1.00/V})\\\\pH=5.25[/tex]
Hence, the pH of the buffer is 5.25