what is the accetic acid solution when 30ml .15M NaOH have been added to 50ml .1M HOAc?
To determine the resulting acetic acid solution after adding 30 ml of 0.15 M NaOH to 50 ml of 0.1 M HOAc, we need to consider the reaction that takes place between NaOH and HOAc.
The balanced equation for the reaction between NaOH and HOAc is:
NaOH + HOAc → NaOAc + H2O
From the balanced equation, we can see that one mole of NaOH reacts with one mole of HOAc to produce one mole of NaOAc and one mole of water.
First, we need to calculate the moles of NaOH and HOAc:
Moles of NaOH = volume (in L) x concentration
Moles of NaOH = 0.030 L x 0.15 mol/L = 0.0045 mol
Moles of HOAc = volume (in L) x concentration
Moles of HOAc = 0.050 L x 0.1 mol/L = 0.005 mol
Next, we need to determine which reactant is limiting. The reactant that is completely consumed will determine the amount of product formed.
From the balanced equation, we can see that the mole ratio between NaOH and HOAc is 1:1. This means that for every mole of NaOH, we need one mole of HOAc to react completely.
Since we have 0.0045 moles of NaOH and 0.005 moles of HOAc, the moles of NaOH is slightly lesser, making it the limiting reactant.
Now, we can determine the moles of acetic acid left after the reaction by subtracting the moles of NaOH consumed from the moles of HOAc initially present.
Moles of HOAc remaining = moles of HOAc initially - moles of NaOH consumed
Moles of HOAc remaining = 0.005 mol - 0.0045 mol = 0.0005 mol
To find the concentration of the acetic acid solution, we need to divide the moles of acetic acid remaining by the volume of the solution:
Concentration of acetic acid solution = Moles of HOAc remaining / Total volume of the solution
Concentration of acetic acid solution = 0.0005 mol / (0.050 L + 0.030 L) = 0.0005 mol / 0.080 L
Concentration of acetic acid solution = 0.00625 mol/L or 0.00625 M
Therefore, the acetic acid solution after adding 30 ml of 0.15 M NaOH to 50 ml of 0.1 M HOAc would have a concentration of approximately 0.00625 M.