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What is the acid ionization constant (ka) of an unknown acid if a 0. 030m solution of it has a ph of 2. 946

Sagot :

The acid ionization constant (ka) of an unknown acid if a 0.030M solution of it has a pH of 2.946 is 4.8×10⁻⁵.

What is the relation between pH and [H⁺]?

Relation between the pH and [H⁺] will be shown as:

[H⁺] = [tex]10^{-pH}[/tex]

Given that pH of solution = 2.946

[H⁺] = [tex]10^{-2.946}[/tex] = 0.00113 = 0.0012

Let the ICE table for the unknown acid HA will be:

                                   HA    ⇄    H⁺   +     A⁻

Initial:                          0.03          0           0

Change:                   -0.0012    0.0012   0.0012

Equilibrium:        0.03-0.0012  0.0012   0.0012

Acid dissociation constant will be calculated as:
Ka = [0.0012][0.0012] / [0.03-0.0012]

Value of 0.0012 is very small as compared to the 0.03, so the equation becomes

Ka = (0.0012)² / 0.03 = 0.000048 = 4.8×10⁻⁵

Hence value of Ka is 4.8×10⁻⁵.

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