# pKb calculation and examples in chemistry?

### pKb in chemistry :

Pkb calculation and examples are listed here. In chemistry pK _{b} is the negative base-10 logarithm of the base dissociation constant (K _{b} ) of a solution . It is used to determine the strength of a base or an alkaline solution.

pKb = -log _{10} K _{b}

The lower the value of pK _{b} , the stronger the base. As for the acid dissociation constant , pK _{a} , the calculation of the basic dissociation constant is an approximation which is only precise in dilute solutions . Kb can be found using the following formula:

K _{b} = [B ^{+} ] [OH ^{–} ] / [BOH]

which is obtained from the chemical equation:

BH ^{+} + OH ^{–} ⇌ B + H _{2} O

### Find pKb of pKa or Ka

The base dissociation constant is related to the acid dissociation constant, so if you know one, you can find the other value. For an aqueous solution, the concentration of hydroxide ions [OH ^{–} follows the relationship of the concentration of hydrogen ions [H ^{+} ] “K _{w} = [H ^{+} ] [OH ^{–}

Putting this relation in the equation K _{b} gives: K _{b} = [HB ^{+} K _{w} / ([B] [H]) = K _{w} / K _{a}

At the same ionic strength and at the same temperatures:

pK _{b} = pK _{w} – pK _{a} .

For aqueous solutions at 25 ° C, pK _{w} = 13.9965 (i.e. around 14), therefore:

pK _{b} = 14 – pK _{a}

### Example of calculation of pK _{b }

Find the value of the base dissociation constant K _{b} and pK _{b} for an aqueous solution at 0.50 dm ^{-3} of a weak base having a pH of 9.5.

First calculate the concentrations of hydrogen and hydroxide ions in the solution to obtain values to plug into the formula.

[H ^{+} ] = 10 ^{-pH} = 10 ^{-9.5} = 3.16 x 10 ^{-10} mol dm ^{-3}

K _{w} = [H ^{+ }_{(aq)} ] [OH ^{– }_{(aq)} ] = 1 x 10 ^{-14} mol ^{2} dm ^{-6}

[OH ^{– }_{(aq)} ] = _{Kw }_{/} [H ^{+ }_{(aq)} ] = 1 x 10 ” ^{14} / 3.16 x 10″ ^{10 }**=** 3.16 x 10 ” ^{5} mole dm” ^{3}

Now you have the information necessary to solve the basic dissociation constant:

K _{b} = [OH ^{– }_{(aq)} ] ^{2 }_{/} [B _{(aq)} ] = (3.16 x 10 ^{-5} ) ** ^{2}** / 0.50

**=**2.00 x 10

^{-9}mol dm

^{-3}

pK _{b} = -log (2.00 x 10 ^{-9} ) **=** 8.70

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