Question

Be sure to answer all parts.

A 0.035 M solution of a weak acid (HA) has a pH of 4.67. What is the K, of the acid?​

Answer 1

To calculate the Ka of a 0.035 M weak acid with pH 4.67, one must convert the pH to [H+], approximate [A-] to [H+], and then use the Ka expression to solve for the acid dissociation constant, resulting in Ka = 1.31 × 10^-8.

Step-by-step explanation:

To find the Ka of the weak acid (HA) using the given pH, we can follow these steps:

1. Convert the pH to [H+]: pH = -log[H+], so [H+] = 10-pH = 10-4.67.
   This gives us [H+] = 2.14 × 10-5 M.
2. For a weak acid, [H+] is approximately equal to the concentration of A-, so [A-] = 2.14 × 10-5 M.
3. Set up the expression for Ka: Ka = [H+][A-]/[HA]. Assuming x is small compared to the initial concentration, [HA] at equilibrium is approximately 0.035 M.
4. Insert the values into the Ka expression: Ka = (2.14 × 10-5)2 / 0.035.
5. Calculate the Ka: Ka = 1.31 × 10-8.

This value of Ka is consistent with that of a weak acid. Typically, two significant figures are appropriate for the answer, since there are two digits after the decimal point in the given pH.

Answer 2

The dissociation constant is 2.44 mM.

Step-by-step explanation:

Dissociation constant of any acid or any kind of solutions is the measure of amount dissolved in the solution or separated into different elements. So it is calculated as the ratio of product of concentration of dissociated elements to the concentration of the original compound.

Here the concentration of the acid [HA] = 0.035 M, the pH of the acid is given as 4.67.

`pH=-log[H^(+)]`

`[H^(+)]=e^(-pH) = (1)/(e^(pH) ) = (1)/(e^(4.67) ) =(1)/(107)`

`[H^(+)]=9.35 * 10^(-3) M`

Then,
`K_(a) =([H^(+)][A^(-)] )/([HA]) =((9.35 * 10^(-3)) ^(2) )/(0.035)\\ \\K_(a) = 2497.78 * 10^(-6)`

Thus, the dissociation constant is 2.44 mM.