Answer:
9.57
Explanation:
Given that:
[tex]pK_{b}=-\log\ K_{b}=-\log(8.0\times 10^{-5})=4.1[/tex]
Considering the Henderson- Hasselbalch equation for the calculation of the pOH of the basic buffer solution as:
[tex]pOH=pK_b+log\frac{[conjugate\ acid]}{[base]}[/tex]
So,
[tex]pOH=4.1+\log\frac{0.540}{0.250}=4.43[/tex]
pH + pOH = 14
So, pH = 14 - 4.43 = 9.57