A 0.180 m solution of a weak acid (ha) has a ph of 2.91. calculate the acid ionization constant (ka) for the acid.

Chemistry

2 answers:

mr_godi [17]3 years ago

8 0

[H+]= [A-]= 10⁻².95=0.00112 M

<span> Ka = [H+][A-]/ [HA]= 0.00112 x 0.00112 / 0.185 =6.81 x 10</span>⁻⁶

N76 [4]3 years ago

3 0

Answer is: <span>the acid ionization constant (Ka) for the acid is </span>8.46·10⁻⁶.
Chemical reaction: HA(aq) ⇄ H⁺(aq) + A⁻(aq).
pH = 2,91.
[H⁺] = 10∧<span>-2,91 = 0,00123 M.
</span>[H⁺] = [A⁻] = 0,00123 M.
[HA] = 0,18 M - 0,00123 M.
[HA] = 0,1787 M.
Ka = [H⁺] · [A⁻] / [HA].
Ka = (0,00123 M)² / 0,1787 M.
Ka = 8,46·10⁻⁶ M.

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