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2 years ago

An unknown weak acid with a concentration of 0.090 M has a pH of 1.80. What is the Ka of the weak acid?

Physics

1 answer:

PolarNik [594]2 years ago

7 0

Answer : The value of $K_a$ of the weak acid is, $3.36\times 10^{-3}$

Explanation : Given,

Initial concentration = 0.090 M

pH = 1.80

First we have to calculate the hydrogen ion concentration.

$pH=-\log [H^+]$

$1.80=-\log [H^+]$

$[H^+]=0.0158M$

Now we have to calculate the $K_a$ of the weak acid.

The dissociation reaction of weak acid is:

$HA\rightleftharpoons H^++A^-$

Initial conc. 0.090 0 0

At eqm. (0.090-x) x x

x = 0.0158 M

The expression for dissociation constant is:

$K_a=\frac{(x)\times (x)}{(0.090-x)}$

Now put all the given values in this expression, we get:

$K_a=\frac{(0.0158)\times (0.0158)}{(0.090-0.0158)}$

$K_a=3.36\times 10^{-3}$

Thus, the value of $K_a$ of the weak acid is, $3.36\times 10^{-3}$

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2 years ago

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2 years ago

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2 years ago

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3 years ago

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I will answer in English.

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