A 0.15 m aqueous solution of the weak base B at 25.0 c has a PH of 8.88 the value of Kb for B is ??

Explanation:

given PH of weak base=8.88

molarity=0.15 m

ionization equation for weak base(B) is

B+H2O⇔HB+OH

kc for this equation is

kc=(HB)(OH)/(B)(H2O)

kc(H2O)=(HB)(OH)/(B)

kc(H2O)=Kb

Kb=(HB)(OH)/(B)

now we have to find the concentration of OH (OH)

we know that

PH+POH=PKw

POH=PKw-PH

POH=14.00-8.88

POH=5.12

POH=log(OH)

(OH)=10^-POH

(OH)=10^-5.12

(OH)=7.58*10^-6

remember that due to molar ratio of 1;1 (HB)=(OH)

so (HB)=(OH)=7.58*10^-6

we will use 0.15 M for (B)

Kb=(HB)(OH)/(B)

Kb=(7.58*10^-6)(7.58*10^-6)/(0.15)

Kb=3.83*10^-10

answer; /// i do believe that the correct answer is (a) boiling

2b2t

explanation:

2b2t