In the titration of 14.0 mL of 0.10M acetic acid with 0.40 M NaOH,
(A) How many mL of NaOH needed to completely neutralize the acid? Given the balanced equation:
CH3COOH (aq) + NaOH (aq) CH3COO Na (aq) + H2O
How many mols acetic acid did you use? That's M x L = ?
Now use the equation coefficients to convert mols acetic acid to mols NaOH. That's a 1:1 ratio; therefore, mols NaOH = same as mols acetic acid.
Then M NaOH = mols NaOH/L NaOH. You have mols and M NaOH, solve for L NaOH and convert to mL.
To determine how many mL of NaOH is needed to completely neutralize the acetic acid, we need to use the balanced equation and stoichiometry.
1. Start by writing the balanced equation:
CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
2. From the balanced equation, we can infer that the ratio of CH3COOH to NaOH is 1:1. This means that for every 1 mole of acetic acid, we need 1 mole of NaOH to neutralize it.
3. Calculate the number of moles of acetic acid (CH3COOH) using the given concentration and volume:
moles of CH3COOH = concentration × volume
moles of CH3COOH = 0.10 M × 0.0140 L
4. Since the ratio of CH3COOH to NaOH is 1:1, the number of moles of NaOH needed is also 0.0140 moles.
5. Now, let's calculate the volume of 0.40 M NaOH required to neutralize the 0.0140 moles of acetic acid. We can use the following formula:
volume of NaOH = moles of NaOH / concentration
volume of NaOH = 0.0140 moles / 0.40 M
6. Calculate the volume of NaOH:
volume of NaOH = 0.0140 moles / 0.40 M
By performing the calculation, you can find the volume of NaOH needed to completely neutralize the acetic acid.