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If the Kb of a weak base is 5.5x10^-6, what is the pH of a 0.38M...

If the Kb of a weak base is 5.5x10-6, what is the pH of a 0.38M solution of this base? 

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Concepts and reason

The concept used is to calculate the of a weak base of the given concentration.

Fundamentals

An equilibrium constant for the dissociation reaction is the equation expressing the extent of dissociation into ions which is equal to the product of the concentrations of the respective ions divided by the concentration of the undissociated molecule.

Consider a general reaction for the dissociation of base, B+H,0
BH* +OH
.

The equilibrium constant for the dissociation of base, is given as follows:

(В)
L-HO ][ Ня)
У

is the measure of hydrogen ion concentration.

The expression to calculate the is as follows:

pH = -log[1,01]

is the measure of hydroxide ion concentration.

The expression to calculate the is as follows:

pOH = -log[OH ]

The hydroxide ion concentration and are related as follows:

[H,0+ ][OH ]=1.0x10-14

The relation between and is as follows:

pH + pOH = 14

Concentration of base [B]=0.38 M

K = 5.5x10

The ICE table is constructed as follows:

B+H20-
Initial: 0.38
Change: -X
Equilibrium: 0.38-x
BH +OH
0 0
+ x + x
x x

The equilibrium concentrations are calculated as follows:

5.5x10- (x)(x)
(0.38-x)
Solve for x.
x=0.00145 M

[BH* )=x=0.00145 M
[OH)= x=0.00145 M
[B]=0.38-x
= 0.38-0.00145 M
= 0.3785 M

The is calculated as follows:

pOH = -log(0.00145 M)
= 2.84

The is calculated as follows:

pH + pOH = 14
pH +2.84 = 14
pH = 14 -2.84
= 11.16

Ans:

Therefore, the of the weak base .

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