TheShiny
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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:
pH = pka + log_{10} \frac {[A^-]}{[HA]} \Rightarrow [H^+] = ka + \frac{[A^-]}{[HA]} \Rightarrow ka = \frac{[H^+][A^-]}{[HA]}
The Attempt at a Solution
Part I
pH = 3.746 \Rightarrow [H^+] = 10^{-3.746} = 1.79473\times10^{-4}
pka = 3.54 \Rightarrow [H^+] = 10^{-3.54} = 2.88403\times10^{-4}
ka = \frac{[H^+][A^-]}{[HA]} \Rightarrow 2.88403\times10^{-4} = 1.79473\times10^{-4}\times\frac{[A^-]}{[HA]}
\therefore \frac{[A^-]}{[HA]} = 1.60694
Pat II
25 mL of 0.244 M acid \Rightarrow 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 |
+---+--------------+--------------+
\frac{[A^-]}{[HA]} = \frac{x}{(0.0061 - x)} \therefore \frac{x}{(0.0061 - x)} = 1.60694
Solve for x...
x = 0.00376009 (is this the moles of acid reacting?)
\frac{0.00376009}{0.275} = 0.01367305
\therefore 13.7 mL NaOH should be added.