# PH of buffer with addition of conjugate base

• Maharg
In summary, the conversation is about finding the pH of a solution containing 0.100 M acetic acid and 0.100 M sodium acetate when 0.0900 mol of sodium acetate is added. The pKa of acetic acid is given as 1.75E-5 and the Henderson-Hasselbach equation is used to solve the problem. The correct approach is to use the equation ka = H+*(0.1+.009)/0.1 and plug the numbers into the HH equation. The final pH value is 9.2E-6.
Maharg

## Homework Statement

Consider 1L of a solution that is 0.100 M in acetic acid and 0.100 M in sodium acetate. What will the pH be, when 0.0900 mol of sodium acetate are added?
Ignore changes in volume.

acetic acid pka = 1.75 E-5

## Homework Equations

HA --> H+ + A-
pH=pka +log A-/HA

## The Attempt at a Solution

I had similar questions that I know how to do with strong acids. The difference in this question is that the added acid/base is a weak one. With the strong base addition I can just add it to A- concentration and subtract it from HA and use Henderson-Hasselbach equation.

This one I tried that but the numbers I know are off. I also tried doing ka = H+*(0.1+.009)/0.1

Solving for H+ and then using that to find out concentrations of A- and HA but the H+ number is far too small to make a difference and I'm fairly sure I can't just add the conjugate bases together.

Maharg said:
I also tried doing ka = H+*(0.1+.009)/0.1

0.009? * instead of log? This is a correct approach, just check your math.

Borek said:
0.009? * instead of log? This is a correct approach, just check your math.

Oops I made it a mistake, anyways when I did my math I just did 1.75E-5 = H+(0.19)/0.1

I get 9.2E-6

but I have no idea what to do with this. If I do it the same way as a strong base question I would add it to the A- concentration and subtract it from HA, but this value is too small to make a difference to a 0.1 M.

You have added conjugate base - so change only base concentration, acid concentration is untouched. Simply plug numbers into HH equation.

## 1. What is a buffer solution?

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added. It is made up of a weak acid and its conjugate base, or a weak base and its conjugate acid.

## 2. How does the addition of a conjugate base affect the pH of a buffer?

The addition of a conjugate base to a buffer solution will cause a slight increase in pH. This is because the conjugate base will react with any added acid, preventing it from significantly changing the pH of the solution.

## 3. What is the Henderson-Hasselbalch equation and how is it related to buffer pH?

The Henderson-Hasselbalch equation is a mathematical expression that relates the pH of a buffer solution to the ratio of the concentrations of the weak acid and its conjugate base. It is used to calculate the pH of a buffer solution before and after the addition of a conjugate base.

## 4. Can a buffer solution have a pH greater than 7?

Yes, a buffer solution can have a pH greater than 7. This is because the pH of a buffer is determined by the equilibrium between the weak acid and its conjugate base, and this equilibrium can exist at any pH value depending on the strength of the acid and base.

## 5. How does the concentration of the conjugate base affect the pH of a buffer solution?

The concentration of the conjugate base in a buffer solution will affect the pH. A higher concentration of conjugate base will result in a higher pH, while a lower concentration will result in a lower pH. This is because the conjugate base plays a key role in maintaining the pH of the buffer solution.

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