The ionization constant of HA is 1 × 10−6
.
What must be the ratio of [A−] to [HA] for
the preparation of a buffer solution with a pH
of 6.77?
whats the answeer??????
Let me start you off, but you have to figure it out from there.
pH = pKa + log ([A-]/[HA])
Solve for ([A-]/[HA]). So you may set it up as
pH = pKa + log x
Now, solve for x...
To determine the ratio of [A-] to [HA] for the preparation of a buffer solution with a pH of 6.77, we can first use the ionization constant (Ka) of HA to calculate the concentration of [H+] in the solution.
The ionization constant is given as Ka = [A-][H+]/[HA], where [A-] represents the concentration of the conjugate base and [HA] represents the concentration of the weak acid.
We can rearrange the equation to solve for [H+], as follows: [H+] = (Ka * [HA]) / [A-].
Given that Ka = 1 × 10^-6, and the pH is 6.77, we can calculate the concentration of [H+] using the equation pH = -log([H+]), which gives [H+] = 10^(-pH).
Substituting these values into the rearranged equation, we have:
10^(-6.77) = (1 × 10^-6 * [HA]) / [A-].
Now, let's assume that the ratio of [A-] to [HA] is x, which means [A-] = x * [HA].
Substituting this into the equation, we get:
10^(-6.77) = (1 × 10^-6 * [HA]) / (x * [HA]).
Simplifying the equation, we have:
10^(-6.77) = 1 / x.
To solve for x, we take the reciprocal of both sides of the equation:
x = 1 / 10^(-6.77).
Evaluating the expression, we have:
x = 10^6.77.
Hence, the ratio of [A-] to [HA] for the preparation of a buffer solution with a pH of 6.77 is 10^6.77.
To answer this question, we need to understand the concept of a buffer solution and how to calculate the ratio of [A-] to [HA] for a given pH.
A buffer solution is a solution that can resist a change in pH when small amounts of acid or base are added to it. It contains a weak acid (HA) and its conjugate base (A-), both in appreciable amounts.
The pH of a buffer solution is determined by the equilibrium between the weak acid and its conjugate base, described by the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where pKa is the negative logarithm (base 10) of the acid dissociation constant (Ka) of the weak acid HA.
In this case, we are given the ionization constant of HA, which is 1 × 10^-6. The ionization constant (Ka) is related to pKa as follows:
Ka = 10^(-pKa)
We can rearrange this equation to solve for pKa:
pKa = -log(Ka)
pKa = -log(1 × 10^-6)
pKa = 6
Now we can substitute the given pH (6.77) and pKa (6) into the Henderson-Hasselbalch equation and solve for the ratio [A-]/[HA]:
6.77 = 6 + log([A-]/[HA])
log([A-]/[HA]) = 6.77 - 6
log([A-]/[HA]) = 0.77
To find the actual ratio [A-]/[HA], we need to convert the logarithm back into a numerical value. We can do this by taking the antilog (inverse logarithm) of both sides:
[A-]/[HA] = antilog(0.77)
Using a calculator, the antilog of 0.77 is approximately 5.89.
Therefore, the ratio of [A-] to [HA] for the preparation of a buffer solution with a pH of 6.77 is approximately 5.89.