titration of a 50.00mL of KOH required 175.00mL of 0.0150M acetic acid, what is the Molarity of KOH solution
CH3COOH + KOH ==> H2O + CH3COOK
mols CH3COOH = M x L = ?
mols KOH = the same (look at the coefficients in the balanced equation.)
M KOH = mols KOH/L KOH.
To find the molarity of the KOH solution, we can use the concept of stoichiometry in a titration. In a titration, a known volume and concentration of one solution (the titrant) is reacted with a known volume and concentration of the other solution (the analyte) until the reaction is complete.
In this case, the known volume and concentration of the analyte (acetic acid) is given as 175.00 mL and 0.0150 M respectively. The volume of the titrant (KOH) is given as 50.00 mL.
The balanced chemical equation for the reaction between acetic acid (CH3COOH) and potassium hydroxide (KOH) is:
CH3COOH + KOH -> CH3COOK + H2O
From the equation, we can see that the molar ratio between acetic acid and KOH is 1:1. This means that 1 mol of acetic acid reacts with 1 mol of KOH.
To determine the number of moles of acetic acid, we can use the formula:
moles = volume x concentration
moles of acetic acid = 0.175 L x 0.0150 mol/L
= 0.002625 mol
Since the molar ratio between acetic acid and KOH is 1:1, the number of moles of KOH is also 0.002625 mol.
Now, we can calculate the molarity of the KOH solution:
molarity = moles / volume
molarity of KOH = 0.002625 mol / 0.050 L
= 0.0525 M
Therefore, the molarity of the KOH solution is 0.0525 M.