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Calculate the [H3O+] of 0.35 M solution of benzoic acid HC7H3O2

Sagot :

sebassandin

Answer:

[tex][H_3O^+]=4.7x10^{-3}M[/tex]

Explanation:

Hello there!

In this case, since the ionization of benzoic acid is:

[tex]HC_7H_3O_2+H_2O\rightleftharpoons C_7H_3O_2^-+H_3O^+[/tex]

Whereas the corresponding equilibrium expression is:

[tex]Ka=\frac{[C_7H_3O_2^-][H_3O^+]}{[HC_7H_3O_2]}[/tex]

Now, we insert the acidic equilibrium constant and the reaction extent x, to write:

[tex]6.3x10^{-5}=\frac{x^2}{0.35M}[/tex]

Thus, by solving for x, we obtain:

[tex]x=\sqrt{6.3x10^{-5}*0.35}\\\\x=4.7x10^{-3}M[/tex]

Which is also:

[tex]x=4.7x10^{-3}M[/tex]

Best regards!

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