80 ml of 1.0M ch3 cooh(aq), are titrated with 2.0M KOH(aq) in the rreaction Ch3cooh(s)+ KOH(aq)reacts with kch3coo(aq)+ H2O what is the volume of potassium hydroxide was used to reach the end point of titration
To find the volume of potassium hydroxide (KOH) used to reach the end point of the titration, we can use the relationship between moles, concentration, and volume.
Let's start by writing down the balanced equation for the reaction:
CH3COOH(aq) + KOH(aq) → CH3COOK(aq) + H2O(l)
From the balanced equation, we can see that the molar ratio between CH3COOH and KOH is 1:1. This means that for every mole of CH3COOH, we need 1 mole of KOH to react completely.
Given that the initial volume of CH3COOH is 80 mL and its concentration is 1.0 M, we can calculate the number of moles of CH3COOH:
Moles of CH3COOH = Volume × Concentration
= 80 mL × 1.0 mol/L
= 80 mmol
Since the molar ratio between CH3COOH and KOH is 1:1, the number of moles of KOH needed to react completely with CH3COOH is also 80 mmol.
Now we need to find the volume of 2.0 M KOH needed to provide 80 mmol of KOH. We can rearrange the formula for moles to solve for volume:
Moles = Volume × Concentration
Volume of KOH = Moles / Concentration
= 80 mmol / 2.0 mol/L
= 40 mL
Therefore, the volume of potassium hydroxide used to reach the end point of the titration is 40 mL.