You have a vinegar solution you believe to be 0.8 M. You are going to titrate 21.39 mL of it with a NaOH solution that you know to be 0.774 M. At what volume of added NaOH solution would you expect to see an end point? Answer in units of mL
Actually, there isn't enough information to answer the question. The below will calculate the volume of NaOH required to reach the EQUIVALENCE POINT (which I assume is what the problem wants). The problem asks, however, mL required to read the END POINT. The end point depends upon the indicator being used and that isn't given in the problem. I would be one value if methyl orange is used, another if methyl red is used, and another if phenolphthalein is used.
Since vinegar is a monoprotic acid and NaOH is monobasic, the ratio is 1:1; therefore, 1 mol acid will be neutralized by 1 mol of the base.
mLacid x M acid = mL base x M base.
Substitute and solve for volume NaOH
Thank You!! I got it right! My answer is 22.10853.
That's a lot of significant figures you are quoting.
To determine the volume of added NaOH solution at which you would expect to see an end point, you can use the stoichiometry of the reaction and the concept of acid-base titration.
The balanced chemical equation for the reaction between acetic acid (the main component of vinegar) and NaOH is:
CH3COOH + NaOH -> CH3COONa + H2O
From the equation, you can see that the ratio between acetic acid and NaOH is 1:1. This means that 1 mole of acetic acid reacts with 1 mole of NaOH.
To determine the moles of acetic acid present in the vinegar solution, you can use the formula:
moles = concentration (M) × volume (L)
Given that the concentration of the vinegar solution is 0.8 M and the volume of vinegar solution used is 21.39 mL (which is equal to 0.02139 L), you can calculate the moles of acetic acid:
moles of acetic acid = 0.8 M × 0.02139 L = 0.017112 mol
Since the stoichiometry between acetic acid and NaOH is 1:1, you would need an equal number of moles of NaOH to react fully with the acetic acid.
Now, you can determine the volume of 0.774 M NaOH solution needed to react with the same number of moles of acetic acid:
moles of NaOH = 0.017112 mol
Using the formula for moles:
moles = concentration (M) × volume (L)
Rearrange the formula to solve for volume:
volume (L) = moles / concentration (M)
Substituting the values:
volume (L) = 0.017112 mol / 0.774 M = 0.022107 L
To convert the volume to milliliters (mL):
volume (mL) = 0.022107 L × 1000 = 22.107 mL
Therefore, you would expect to see an end point at approximately 22.107 mL of added NaOH solution.