211k views
2 votes
Calculate ka for weak acid HA which has a pH of 3.65 at 0.30 concentration.​

2 Answers

4 votes

Answer:

the Key value for the weak acid HA is 1.18 x 10^(-7).

Step-by-step explanation:

The pH of a weak acid solution can be related to the dissociation constant, Key, of the acid through the following equation:

pH = pKa + log([A-]/[HA])

where pH is the measured pH of the solution, pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base of the acid, and [HA] is the concentration of the UN dissociated acid.

In this problem, we are given the pH and the concentration of the weak acid, HA. We need to find the value of Ka.

First, we can rearrange the above equation to solve for pKa:

pKa = pH - log([A-]/[HA])

Next, we need to find the concentration of the conjugate base, [A-]. Since HA is a weak acid, we can assume that only a small fraction of it has dissociated into A- and H+ ions. At equilibrium, the concentration of A- can be assumed to be equal to the concentration of H+ ions produced by the dissociation of HA. We can calculate this concentration from the pH:

[H+] = 10^(-pH) = 10^(-3.65) = 2.24 x 10^(-4) M

Therefore, [A-] = 2.24 x 10^(-4) M.

We can now substitute the values we have calculated into the equation for pKa to obtain:

pKa = 3.65 - log(2.24 x 10^(-4)/0.30) = 3.65 + 2.47 = 6.12

Finally, we can calculate the Ka value from the pKa:

Ka = 10^(-pKa) = 10^(-6.12) = 1.18 x 10^(-7)

Therefore, the Ka value for the weak acid HA is 1.18 x 10^(-7).

User Artem
by
7.9k points
3 votes

Answer:

the Ka value for the weak acid HA is 1.18 x 10^(-7).

Step-by-step explanation:

The pH of a weak acid solution can be related to the dissociation constant, Ka, of the acid through the following equation:

pH = pKa + log([A-]/[HA])

where pH is the measured pH of the solution, pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base of the acid, and [HA] is the concentration of the undissociated acid.

In this problem, we are given the pH and the concentration of the weak acid, HA. We need to find the value of Ka.

First, we can rearrange the above equation to solve for pKa:

pKa = pH - log([A-]/[HA])

Next, we need to find the concentration of the conjugate base, [A-]. Since HA is a weak acid, we can assume that only a small fraction of it has dissociated into A- and H+ ions. At equilibrium, the concentration of A- can be assumed to be equal to the concentration of H+ ions produced by the dissociation of HA. We can calculate this concentration from the pH:

[H+] = 10^(-pH) = 10^(-3.65) = 2.24 x 10^(-4) M

Therefore, [A-] = 2.24 x 10^(-4) M.

We can now substitute the values we have calculated into the equation for pKa to obtain:

pKa = 3.65 - log(2.24 x 10^(-4)/0.30) = 3.65 + 2.47 = 6.12

Finally, we can calculate the Ka value from the pKa:

Ka = 10^(-pKa) = 10^(-6.12) = 1.18 x 10^(-7)

Therefore, the Ka value for the weak acid HA is 1.18 x 10^(-7).

User Aditya Nikhade
by
7.1k points