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Vladimir [108]
3 years ago
8

A 0.15 m solution of chloroacetic acid has a ph of 1.86. What is the value of ka for this acid?

Chemistry
1 answer:
dem82 [27]3 years ago
4 0

Answer: 1.67\times 10^{-3}

Explanation:

ClCH_2COOH\rightarrow ClCH_2COO^-+H^+

   cM              0             0

c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

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

Given:  c = 0.15 M

pH = 1.86

K_a = ?

Putting in the values we get:

Also pH=-log[H^+]

1.86=-log[H^+]

[H^+]=0.01

[H^+]=c\times \alpha

0.01=0.15\times \alpha

\alpha=0.06

As [H^+]=[ClCH_2COO^-]=0.01

K_a=\frac{(0.01)^2}{(0.15-0.15\times 0.06)}

K_a=1.67\times 10^{-3]

Thus the vale of K_a for the acid is 1.67\times 10^{-3}

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We need to convert 1.2\times 10^{-8}\ s to nanoseconds.

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