a solution of a weak monoprotic base b(0.50M)and its salt BH+Cl-(.80M) has a ph=9.40. Need value of Kb 4 baseB

Answer 1

It's a buffer solution
At equilibrium the base concentration is 0.5-x, the salt concentration is 0.80+x and the OH- concentration is x.
pH+pOH=14
pOH=14-pH=4.6
OH- =0.0000251 M
K= (0.8+0.0000251)(0.0000251/ /0.5-0.0000251 = 0.0000402=
= 4.02 10^-5

Answer 2

In such cases you usually use the Henderson-Hasselbalch equation

pH=pKa+log[base]/[acid]
Here the weak acid is BH+ and the conjugate base B
we also know that Ka=Kw/Kb => pKa=pKw-pKb=14-pKb

so

pH=14-pKb +log[B]/[BH+] =>
pKb= 14-pH +log[B]/[BH+] =14-9.40+log(0.5/0.8) =4.396 =>
Kb=10^-4.396 =4.02*10^-5