# Adding a strong acid to buffer solution

• kevvyroo
In summary, the student is seeking assistance in finding the pH change of adding 0.001 mol of HNO3 to a buffer solution. They have provided relevant equations and have attempted a solution, but their final answer does not match the expected value. Further calculation using more accurate values may help to find the correct answer.
kevvyroo

## Homework Statement

I'm trying to find the pH change of adding 0.001 mol of HNO$_{}3$ to a buffer solution containing 0.24 M HF and 0.55 M NaF in 0.100 L.

K$_{}a$ = 3.5E-4 therefore pK$_{}a$=3.47

## Homework Equations

pH=pK$_{}a$+log$_{}10$[base]/[acid]
(Henderson-Hasselbach equation)
and of course
-log$_{}10$(H$^{}+$)=pH

## The Attempt at a Solution

I've tried an ICE table but my best guess for a solution was
0.01 M H+ (from HNO$_{}3$) react with the F$^{}-$ base ions to form HF.
.55 M( fluoride ions)-.01 M( hydronium)=.54 (final fluoride ions)
0.24 M (HF) + 0.01 M (HF produced from above reaction) = 0.25 M HF
So,
pH=3.47+log$_{}10$(.54/.25)
pH=3.80

The answer that is supposed to be correct is 3.79... is this just a sig figs error or did I mess up along the line because I needed exactly 3.79 to be the pH

Sorry if I did anything wrong, I'm new to this forum! (:

Don't use LaTeX just for subscripts - it doesn't work. It is better to either enter the expression entirely as LaTeX - so pK_a will look like $pK_a$ or to format it using [noparse][/noparse] and/or [noparse][/noparse] tags - so [noparse]HNO3[/noparse] will become HNO3.

Your approach is correct, just 3.47 is not correct enough. Try to calculate

$$-log(3.5\times 10^{-4})+log(\frac {0.54}{0.25})$$

in one step, or at least without rounding down intermediate values.

Note, that 3.79 doesn't make much sense - neither concentrations nor Ka are given with accuracy high enough to justify giving answer with three significant digits. We often report calculation of pH with two decimal digits, but we rarely should.

## 1. What is a buffer solution?

A buffer solution is a solution that is able to resist changes in pH when a strong acid or base is added to it. It typically consists of a weak acid and its conjugate base, or a weak base and its conjugate acid.

## 2. How does a buffer solution work?

A buffer solution works by utilizing a process called the Henderson-Hasselbalch equation, which maintains a relatively constant pH by balancing the concentrations of the weak acid and its conjugate base. When a strong acid is added, the weak acid component of the buffer will neutralize it, preventing a drastic change in pH.

## 3. What happens when a strong acid is added to a buffer solution?

When a strong acid is added to a buffer solution, the weak acid component of the buffer will neutralize the acid, preventing a significant decrease in pH. The conjugate base of the weak acid will also be formed, which helps to further resist changes in pH.

## 4. Can a buffer solution handle any amount of strong acid?

No, a buffer solution has a limited capacity to neutralize strong acids. Once the weak acid component is used up, the pH will start to decrease rapidly as more strong acid is added. It is important to choose a buffer with an appropriate concentration and pH range for the desired application.

## 5. How can I prepare a buffer solution to add a strong acid?

To prepare a buffer solution, you will need to choose a weak acid and its conjugate base or a weak base and its conjugate acid. You will also need to determine the desired pH range and concentration of the buffer. Then, you can mix the components in the appropriate ratios to achieve the desired buffer solution. Alternatively, pre-made buffer solutions can also be purchased.

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