Question

For a certain acid pKa =7.26. Calculate the pH at which an aqueous solution of this acid would be 0.076% dissociated. Round your answer to 2 decimal places

Answer 1

4.25

Step-by-step explanation:

To solve this, we need to use the dissociation constant (Ka) of the acid, which is related to its pKa value.

The dissociation constant is the ratio of the concentrations of the dissociated and undissociated forms of the acid. For a weak acid like this one, we can assume that the concentration of the undissociated form is approximately equal to the total concentration of the acid.

The equation for the dissociation constant (Ka) is:

Ka = [H+][A-] / [HA]

where [H+] is the concentration of hydrogen ions, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the undissociated acid.

We can rearrange this equation to solve for the hydrogen ion concentration:

[H+] = sqrt(Ka*[HA]/[A-])

where sqrt means "square root".

Now, we need to use the percent dissociation to find the concentration of the undissociated acid ([HA]) and the concentration of the dissociated form ([A-]).

If the acid is 0.076% dissociated, that means that 99.924% of the acid is still in its undissociated form. So:

[HA] = 0.99924 * initial concentration of acid

[A-] = 0.00076 * initial concentration of acid

Now we can plug these values into the equation for [H+] that we derived earlier:

[H+] = sqrt(Ka*[HA]/[A-])

[H+] = sqrt(10^-7.26 * 0.99924 / 0.00076)

When we solve for [H+], we get:

[H+] = 5.63 * 10^-5

Finally, we can use the definition of pH (pH = -log[H+]) to find the pH of the solution:

pH = -log(5.63 * 10^-5)

pH = 4.25

So the pH at which an aqueous solution of this acid would be 0.076% dissociated is 4.25 (rounded to 2 decimal places).

Answer 2

pH ≈ 7.26 - (-3.12) ≈ 10.38

Step-by-step explanation:

To calculate the pH at which an aqueous solution of the acid is 0.076% dissociated, you can use the formula for percent dissociation:

Percent Dissociation = (Dissociated / Initial concentration) * 100

In this case, the percent dissociation is given as 0.076%. Let's denote the initial concentration of the acid as [A].

0.076% = (Dissociated / [A]) * 100

To simplify the equation, divide both sides by 100:

0.00076 = Dissociated / [A]

Now, let's assume that the initial concentration [A] is equal to the concentration of the undissociated acid [HA] since only a small fraction dissociates.

0.00076 = Dissociated / [HA]

The dissociated portion can be expressed as [H+] since it represents the concentration of hydrogen ions in the solution. Since the acid is monoprotic, the concentration of [A-] (conjugate base) is equal to [H+].

0.00076 = [H+] / [HA]

To solve for [H+], take the negative logarithm (base 10) of both sides:

-log([H+]) = -log([HA]) + log(0.00076)

Since pKa = -log(Ka), where Ka is the acid dissociation constant, we can rewrite the equation as:

-pH = -pKa + log(0.00076)

Rearranging the equation, we get:

pH = pKa - log(0.00076)

Substituting the given pKa value (7.26) into the equation:

pH = 7.26 - log(0.00076)

Now, calculate the value using a calculator:

pH ≈ 7.26 - (-3.12) ≈ 10.38

Therefore, at a pH of approximately 10.38, the aqueous solution of the acid would be 0.076% dissociated.