Question & Answer: If a buffer solution is 0.190 M in a weak base (K_b = 8.2 times 10^-5) and 0.440 M in its…..

Kb = 8.2*10^-5

pKb = – log (Kb)

= – log(8.2*10^-5)

= 4.0862

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.0862+ log {0.44/0.19}

= 4.45

use:

PH = 14 – pOH

= 14 – 4.45

= 9.55

Answer: 9.55