Question

What is the Kb of X1- if a 0.0220-M solution of KX has a pH of 12.007?

Kb

Answer 1

Answer:- Kb of X- = 8.72 x 10-3

Solution:-

It is given that

pH = 12.007

Since we know that

pOH = 14 - pH

or, pOH = 14 - 12.007

or, pOH = 1.993

or, - log[OH-] = 1.993

or, log[OH-] = - 1.993

or, [OH-] = 10-1.993

or, [OH-] = 0.0101625

Since we know that

KX + H2O $1586071384769_blob.png$ XH + OH-

at equilibrium, [OH-] = 0.0101625 M = [XH]

Initial concentration of [KX] = 0.0220 M

Hence, the equilibrium concentration of [KX] = 0.0220 M - 0.0101625 M = 0.0118375 M

Then we know that

Kb = [XH][OH-]/[KX]

or, Kb = (0.0101625)( 0.0101625)/ 0.0118375

or,  Kb = 0.00872 = 8.72 x 10-3

Hence, Kb of X- = 8.72 x 10-3 Answer (answer must have three significant figures)

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