Find moles of NaOH required(used) to react with CH3COOH.
The volume of NaOH used was 7.70mL and the actual molarity of NaOH solution is .275 mol/L.
you don't say how much CH3COOH is involved.
So who knows how much NaOH will be needed?
But, the amount of NaOH present is
7.70mL * 1L/1000mL * 0.275 mol/L = 0.00212 mol
To find the moles of NaOH required to react with CH3COOH, you need to know the balanced chemical equation for the reaction between NaOH and CH3COOH. Assuming that the reaction is a 1:1 ratio between NaOH and CH3COOH, the balanced equation is:
NaOH + CH3COOH → CH3COONa + H2O
From the equation, you can see that one mole of NaOH reacts with one mole of CH3COOH.
Given:
Volume of NaOH used (V) = 7.70 mL = 0.00770 L
Molarity of NaOH solution (M) = 0.275 mol/L
To calculate the moles of NaOH used, you can use the formula:
Moles of NaOH (n) = Molarity (M) × Volume (V)
Substituting the given values:
Moles of NaOH (n) = 0.275 mol/L × 0.00770 L
Calculating the moles of NaOH:
n = 0.0021175 mol
Therefore, the moles of NaOH required (used) to react with CH3COOH is approximately 0.0021 mol.