What volume of 0.050M sodium hydroxide is needed to react with 25cm3 of 0.068M ethanoic acid?
To solve this problem, we can use the concept of stoichiometry.
First, let's write down the balanced chemical equation for the reaction between sodium hydroxide (NaOH) and ethanoic acid (CH3COOH):
CH3COOH + NaOH -> CH3COONa + H2O
From the balanced equation, we can see that one mole of sodium hydroxide reacts with one mole of ethanoic acid.
Now, let's calculate the number of moles of ethanoic acid in 25 cm³ of 0.068 M ethanoic acid solution:
Moles of ethanoic acid = volume (in liters) × concentration
Moles of ethanoic acid = 25 cm³ × (1 L / 1000 cm³) × 0.068 M
Next, we need to determine the volume of sodium hydroxide required to react with the calculated moles of ethanoic acid. Since the stoichiometry tells us that the ratio of NaOH to CH3COOH is 1:1, the number of moles of sodium hydroxide needed will be the same as the moles of ethanoic acid.
Finally, we can calculate the volume of 0.050 M sodium hydroxide solution required to react with the determined moles of ethanoic acid:
Volume (in liters) = moles ÷ concentration
Volume (in liters) = moles ÷ 0.050 M
To summarize:
1. Calculate the moles of ethanoic acid using the given volume and concentration.
2. Use the moles of ethanoic acid to determine the moles of sodium hydroxide required.
3. Calculate the volume of 0.050 M sodium hydroxide solution using the moles of sodium hydroxide and its concentration.
Please provide the moles of ethanoic acid calculated in step 1 in order to proceed with the final calculation.