Answer:
7.37 is the pH of the buffer.
Explanation:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=pK_a+\log(\frac{[Na_2HPO_4]}{[NaH_2PO_4]})[/tex]
We are given:
[tex]K_a[/tex] = dissociation constant = [tex]K_a=6.2\times 10^{-8}[/tex]
Moles of [tex]Na_2HPO_4=\frac{20.1 g}{141.95884056 g/mol}=0.1416 mol[/tex]
Moles of [tex]NaH_2PO_4=\frac{11.6 g}{119.97701128 g/mol}=0.09668 mol[/tex]
[tex][concentration]=\frac{moles}{volume (L)}[/tex]
[tex][NaH_2PO_4]=\frac{0.09668 mol}{1 L}=0.09668 M[/tex] (acid)
[tex][Na_2HPO_4]=\frac{0.1416 mol}{1 L}=0.1416 M[/tex] (salt)
pH = ?
Putting values in above equation, we get:
[tex]pH=-\log[6.2\times 10^{-8}]+\log(\frac{0.1416 M}{0.09668 M})\\\\pH=7.37[/tex]
7.37 is the pH of the buffer.
