A buffer solution is made using a weak acid, HA , that has a p𝐾a of 5 . If the pH of the buffer is 7 , what is the ratio of [A−] to [HA] ? [A−][HA]=

Answer 1

Bottom Line (but I put it at the top so you don't need to scroll for knowledge):

Every time we decrease pH by 1 (get closer to acidic end), the molar concentration of acid increases 10x.

Every time we increase pH by 1 (get closer to basic end), the molar concentration of acid decreases 10x.

pH is shorthand for the Power of Hydrogen. When you think of pH, the numbers 1-14 pop into your mind, with 1 being the "acidic" side of the spectrum, and 14 being the "basic" side.

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It is important to know that every time we increase or decrease our pH value by 1, we subsequently increase or decrease the concentration of our acid or base 10x!!

This is because pH is calculated on a logarithmic scale.

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So, you might be thinking... what does that mean practically?

Well, take for example Beaker A with a solution with pH 5, and Beaker B with a solution of pH 6. Even though the pH number only increases by 1 between A and B, A is 10x more acidic than B.

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So - let's answer the question at hand.

Beaker A is our weak acid (pH 6)

Beaker B is our buffer solution (pH 8)

We know that increasing our pH by 1 = a 10x decrease in molar concentration of acid since pH is not linear, it is logarithmic.

So, increasing our pH by 2 does NOT equal a 20x decrease in molar concentration of acid. It equals a 100x decrease!!

How? The answer lies with exponents.

The equation to remember is:

10^{(pH of final) - (pH of initial)}

If we take our problem, we get 10^(8)-(6) = 10² = 100

Let's break this down to solve the problem:

[A-] = base/buffer

[HA] = weak acid

Let's 'set' the base to have a ratio of 1, meaning that at pH of 8, there is 1 molar equivalent of base present.

The weak acid must meet the pH of 8 where it is, as this is our 'setpoint'. For the weak acid to 'meet' the base where it is, the weak acid must increase twice. First to 7, then to 8.

Since it INCREASED twice, it had to decrease in molar concentration by 10² = 100.

Hence, our ratio of weak acid [HA] = 1 / 100 = 0.01 WHEN our base [A-] = 1. In other words, the weak acid is 1/100th the concentration of the base, and the base is 100x as concentrated as the acid.

[A-] = 1

[HA] = 0.01

we can multiply both concentrations by 100 to relieve the acid of its decimal:

[A-] = 100

[HA] = 1

Hooray! That is our answer. Please reach out with any questions or comments about this answer. I'd love your feedback. Good luck, you got this.

Answer 2

For this problem we will be using the Henderson–Hasselbalch equation, which states:

pH = pKa + log₁₀([A⁻]/[HA])

We know the pH of our solution is 7, and the pKa of our weak acid is 5. So we can plug in these values and solve for the ratio [A⁻]/[HA].

7 = 5 + log₁₀([A⁻]/[HA])

2 = log₁₀([A⁻]/[HA])

10² = [A⁻]/[HA]

100 = [A⁻]/[HA]