Question

33. A monoprotic weak acid when dissolved in water is 0.36% dissociated and produces a solution...

33. A monoprotic weak acid when dissolved in water is 0.36% dissociated and produces a solution with a pH of 3.10. Calculate the Ka of the acid. SHOW WORK!!!

A) 3.6 × 10–3

B) 2.2 × 10–1

C) 2.9 × 10–6

D) Need to know the initial concentration of the acid.

E) None of these.

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Answer #1

If it is a Strong acid it will dissociate completely.

If we are considering this as a strong acid HA ----> it will dissociate is water as H+ and A- equally, as we can say 100%

If it is a week acid the dissociation will be very less. Some of the ions will dissociate but equally and remaining won't dissociate.

So here we have taken week acid. Some will dissociate as H+ and A- equally. So the concentration of H+ and A- will be in equal. The initial concentration of HA will be high. After dissociation, the concentration of HA will be decreased because some of the HA week acids dissociates into H+ and A-.

The formula for finding Ka is

Here [H+] = Concentration of H+ after the dissociation

[A-] = Concentration of A- after the dissociation

[HA] = Concentration of HA after the dissociation.

Finding the concentration of [H+] and [A-]

Always [H+] and [A-] will be equal. [H+] = [A-]

we know that we can calculate the pH for the dissociated ions. Here the pH is 3.1 so which refers to the pH of [H+]

So, pH = -log [H+]

[H+] = pH/ -log

[H+] = -pH/ log that we can write it as [H+] = 10-pH

[H+] = 10-pH

pH= 3.1 (given in the problem)

[H+] = 10-3.1

[H+] = 0.00079 same as for [A-] = 0.00079

Calculate the Concentration of [HA] after the dissociation

If 0.36% of HA dissociates, that means 0.36% of [HA] is dissociated after added into water. of its initial concentration of [HA].  

This 0.36% we can write as 0.36/100. because % is mentioning that whatever out of 100. so 0.0036 this much has dissociated.

We know the Concetration of dissociation which is nothing but, [H+] = 0.00079

so,

0.00079 = 0.0036 X initial concentration

initial concentration = 0.00079 / 0.0036

initial concentration = 0.2194

As in the formula [HA] = Concentration of HA after the dissociation. To calculate this we need to reduce the concentration of dissociated ions [H+] from the initial concentration of [HA]

[HA] = Initial concentration (0.2194) - Dissociated concentration (0.00079]

= 0.2194 - 0.00079

[HA] = 0.21861

To substitute all the values in the euation,

[H+] = 0.00079

[A-] = 0.00079

[HA] = 0.21861

Ka = 2.85 X 10-6

Round off the digit, we will get 2.9 X 10-6. Because the second digit after the decimal is equal to 5 so we can add 1 to the first digit after the decimal.

Ka = 2.9 x 10-6 Option C is the correct value.

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