Assume the vinegar is 5.00% acetic acid by mass and the density of vinegar is 1.00 g/mL. Assume you have a 0.100 M NaOH solution. Calculate the volume (in mL) of NaOH needed to titrate 5.00 mL of vinegar.

Question

matt101 · Answer

You have 5 mL of vinegar. The density of vinegar is 1 g/mL, so the mass is:

$$density={mass \over volume}$$$$1={mass \over 5}$$$$mass =5$$
So there are 5 g of vinegar in 5 mL of vinegar. We know that 5% of the mass of vinegar is acetic acid, CH3COOH (the other 95% is just water). 5% of 5 g is 0.05 x 5 = 0.25 g of CH3COOH. The balanced neutralization reaction for CH3COOH with NaOH is:

$$CH_3COOH + NaOH \rightarrow H_2O+CH_3COONa$$
You can see from this reaction that the ratio of CH3COOH to NaOH is 1:1, so equal moles of both must react. If we know how many moles of CH3COOH we have, we'll know how many moles of NaOH were used. We can find moles of CH3COOH by dividing its mass by its molar mass:

$$n={m \over MM}$$$$n={0.25 \over 60}$$$$n=4.17 \times 10^{-3}$$
There are 4.17 x 10^(-3) mol of CH3COOH in your solution, so you need just as many moles of NaOH to neutralize it. We have the concentration of our NaOH solution, so we can use that to find the volume required:

$$C={n \over V}$$$$0.1={4.17 \times 10^{-3} \over V}$$$$V=4.17 \times 10^{-2}$$
So you need 4.17 x 10^(-2) L of NaOH solution. In mL (multiply by 1000), this is 41.7 mL.

If you have any questions please let me know!