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FrozenT [24]
3 years ago
13

The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.

Round your answer to 1 decimal place.
Chemistry
1 answer:
madam [21]3 years ago
3 0

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

H_3BO_3\rightarrow H^+H_2BO_3^-

at t=0  cM              0             0

at eqm c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 4.4 M and \alpha = ?

K_a=5.8\times 10^{-10}

Putting in the values we get:

5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}

(\alpha)=0.000011

[H^+]=c\times \alpha

[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M

Also pH=-log[H^+]

pH=-log[4.8\times 10^{-5}]=4.3

Thus pH of a 4.4 M H_3BO_3 solution is 4.3

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