Solution :
[tex]pk_b = -log ( K_b)\\\\pk_b = - log ( 3.6 \times 10^{-5})\\\\pk_b = 4.44[/tex]
We know, pH of this solution :
[tex]pH = pK_a + log ( \dfrac{|conjugate\ base|}{|weak \ acid|}\\\\pH = 4.44 + log(\dfrac{0.480}{0.190})\\\\pH = 4.44 + 0.40\\\\pH =4.84[/tex]
Hence, this is the required solution.