Question

Suppose you need to prepare 250.0 mL of a pH=3.60 buffer that has a total buffer concentration of formic acid + formate of 0.030 M. How many moles of formic acid will you need? The pKa of formic acid is 3.75.

Answer 1

Answer:

The number of moles of formic acid needed is 4.5x10⁻³ moles.

       

Explanation:

We can find the moles of formic acid using the Henderson-Hasselbalch equation:

$pH = pKa + log(\frac{[CHOO^{-}]}{[CHOOH]})$

We know:

pH = 3.60

pKa = 3.75                    

By solving the above equation for [CHOO⁻]/[CHOOH] we have:

$\frac{[CHOO^{-}]}{[CHOOH]} = 10^{(pH - pKa)} = 10^{(3.60 - 3.75)} = 0.71$

$[CHOO^{-}] = 0.71[CHOOH]$  (1)

Now, we have:

$[CHOOH] + [CHOO^{-}] = 0.03 M$   (2)

By entering equation (1) into (2) we have:

$[CHOOH] + 0.71[CHOOH] = 0.03 M$

$[CHOOH] = 0.018 M$

Hence, the concentration of formate is:

$[CHOO^{-}] = (0.03 - 0.018)M = 0.012 M$

Finally, the number of moles of formic acid is:

$n_{CHOOH} = [CHOOH]*V = 0.018 \frac{mol}{L}*0.250 L = 4.5 \cdot 10^{-3} moles$

Therefore, 4.5x10⁻³ moles of formic acid are needed.

I hope it helps you!  

Answer 2

We have that for the Question "How many moles of formic acid will you need?"

It can be said that

• $4.39*10^{-3}$ moles of formic acid will be needed

From the question we are told

to prepare 250.0 mL of a pH=3.60, total buffer concentration of formic acid + formate of 0.030 M, The pKa of formic acid is 3.75.

Let $1+COO4 = X, 1+COON = (0.03-X)$

$pH = pKa + log\frac{1+COON}{1+COO4}\\\\3.6 = 3.75 +log\frac{0.03-X}{X}\\\\log\frac{0.03-X}{X} = 3.6 - 3.75\\\\\frac{0.03-X}{X} = 0.708\\\\X = 0.0175M$

Therefore,

$moles of formic acid = 0.0175*0.25mole\\\\= 4.39*10^{-3}mole$

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