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3 years ago

The Ka of a monoprotic weak acid is 0.00732. What is the percent ionization of a 0.123 M solution of this acid?

Chemistry

1 answer:

MaRussiya [10]3 years ago

5 0

Answer:

21.60% is the percent ionization of a 0.123 M solution of this acid.

Explanation:

The equilibrium reaction for dissociation of weak acidis,

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

initially conc. c 0 0

At eqm. $c(1-\alpha )$ $c\alpha$ $c\alpha$

Concentration of the weak acid (c) = 0.123 M

Acid dissociation constant = $k_a=0.00732$

Degree of ionization of weak acid = $\alpha$

An expression of dissociation constant is given as:

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha )}=\frac{c\times (\alpha )^2}{(1-\alpha )}$

$0.00732=\frac{0.123 M\times (\alpha )^2}{(1-\alpha )}$

$\alpha =0.2160$

Percent ionization of weak acid:

$\% ionization=\frac{c\alpha }{c}\times 100$

$\frac{0.123 M\times 0.2160}{0.123 M}\times 100=21.60\%$

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2 years ago

Malonic acid, h2c3h2o4, is a diprotic acid with ka1= 1.5 x 10-3and ka2= 2.0 x 10-6. what is the concentration of the malonate an

Bess [88]

In a 0.15 m solution of malonic acid, the amount of the malonate anion, c3h2o42-, is $1.68*10^{-4} M$ .

Explanation:

For the first equlibria;

initial concentration of $H_{2}C_{3}H_{2}OH=0.15$

initial concentration of $HC_{3}H_{2}O_{4}^{-} =0$

initial concentration of $H^{+} =0$

Let concentration change of $H_{2}C_{3}H_{2}OH=-x$

Let concentration change of $HC_{3}H_{2}O_{4}^{-} =+x$

Let concentration change of $H^{+} =+x$

Let concentration in equilibrium state of $H_{2}C_{3}H_{2}OH=0.15-x$

Let concentration in equilibrium state of $HC_{3}H_{2}O_{4}^{-} =x$

Let concentration in equilibrium state of $H^{+} =x$

Therefore, $Ka_{1}=\frac{x*x}{0.15-x}=\frac{1.5*10^{-3}}{1} =\frac{x^{2}}{0.15-x}$

$x^{2}+0.0015x-0.000225=0$

Which is a quadratic equation in x, solving it gives;

$x=0.01427$

Therefore, $HC_{3}H_{2}O_{4}^{-} =x= 0.01427M$

Now for the second equlibria;

initial concentration of $HC_{3}H_{2}O_{4}^{-}=0.01427$

initial concentration of $C_{3}H_{2}O_{4}^{2-} =0$

initial concentration of $H^{+} =0$

Let concentration change of $HC_{3}H_{2}O_{4}^{-}=-x$

Let concentration change of $C_{3}H_{2}O_{4}^{2-} =+x$

Let concentration change of $H^{+} =+x$

Let concentration in equilibrium state of $HC_{3}H_{2}O_{4}^{-}=0.01427-x$

Let concentration in equilibrium state of $C_{3}H_{2}O_{4}^{2-} =x$

Let concentration in equilibrium state of $H^{+} =x$

Therefore, $Ka_{2}=\frac{x*x}{0.01427-x}=\frac{2.0*10^{-6}}{1} =\frac{x^{2}}{0.01427-x}$

$2.854*10^{-8}-(2.0*10^{-6})x=x^{2}$

Therefore, $C_{3}H_{2}O_{4}^{2-} =x= 1.68*10^{-4}M$

<h3>What is a malonic acid?</h3>

The chemical formula of malonic acid is CH2(COOH)2. Malonates include the ionized form of malonic acid as well as its esters and salts. For instance, the diethyl ester of malonic acid is called diethyl malonate.

Polyester and polymers both employ malonic acid as a precursor. Both the scent business and the flavoring sector utilize it. It is employed to regulate acidity. Malonic acid contributes to the production of fatty acids in people. Malonic acid has been found in a number of foods outside of the human body, including red beets, corn, scarlet beans, common beets, and cow milks, although its presence has not been defined. The group of organic substances known as dicarboxylic acids and derivatives includes malonic acid, also known as malonate or H2MALO.

To know more about malonic acid, visit:

brainly.com/question/13943912?referrer=searchResults

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4 0

1 year ago