Answer:
pH of the solution = 4.80
Explanation:
pKa = 8.3
Concentration of tris acid = 8.3
[tex]pH = \frac{1}{2} \times p_{ka} - \frac{1}{2} logC[/tex]
[tex]pH = \frac{1}{2} \times 8.3 - \frac{1}{2}\times log(0.05)[/tex]
[tex]= \frac{1}{2} \times 8.3 - \frac{1}{2}\times log(5\times 10^{-2})[/tex]
[tex]= 4.15 - \frac{1}{2}\times (-1.30103)[/tex]
[tex]= 4.15 - (-0.650515)[/tex]
[tex]= 4.15 + 0.650515[/tex]
= 4.80
pH of the solution = 4.80
