- #1

TheShiny

- 9

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## Homework Statement

Prepare a buffer solution of pH 3.746 from 25 mL of 0.244 M weak acid with pka 3.54

What volume, in mL, of 0.275 M NaOH would need to be added?

I think that I have the answer but I am not sure, could you look over the working to check?

Also, would you do this differently? If so, what would you do?

## Homework Equations

Henderson-Hasselbach:

[itex]pH = pka + log_{10} \frac {[A^-]}{[HA]} \Rightarrow [H^+] = ka + \frac{[A^-]}{[HA]} \Rightarrow ka = \frac{[H^+][A^-]}{[HA]}[/itex]

## The Attempt at a Solution

__Part I__

[itex]pH = 3.746 \Rightarrow [H^+] = 10^{-3.746} = 1.79473\times10^{-4}[/itex]

[itex]pka = 3.54 \Rightarrow [H^+] = 10^{-3.54} = 2.88403\times10^{-4}[/itex]

[itex]ka = \frac{[H^+][A^-]}{[HA]} \Rightarrow 2.88403\times10^{-4} = 1.79473\times10^{-4}\times\frac{[A^-]}{[HA]}[/itex]

[itex]\therefore \frac{[A^-]}{[HA]} = 1.60694[/itex]

__Pat II__

25 mL of 0.244 M acid [itex]\Rightarrow[/itex] 0.0061 M (6.1 mM)

Code:

```
+---+--------------+--------------+
| | Acid | Base |
| I | 0.0061 M | 0 M |
| C | -x M | +x M |
| E | (0.0061-x)M | x M |
+---+--------------+--------------+
```

[itex]\frac{[A^-]}{[HA]} = \frac{x}{(0.0061 - x)} \therefore \frac{x}{(0.0061 - x)} = 1.60694[/itex]

Solve for [itex]x[/itex]...

[itex]x = 0.00376009[/itex] (is this the moles of acid reacting?)

[itex]\frac{0.00376009}{0.275} = 0.01367305[/itex]

[itex]\therefore[/itex] 13.7 mL NaOH should be added.