A 14.7 mL sample of 0.0550 M KOH solution required 55.5 mL of aqueous acetic acid solution in a titration experiment. Calculate the molarity of the acetic acid solution.
HAc = acetic acid
HAc + KOH ==> KAc + H2O
mols KOH = M x L = ?
Convert mols KOH to mols HAc. Since 1 mol HAc = 1 mol KOH (from the coefficients), then mols HAc = mols KOH.
Then M HAc = mols HAc/L HAc = ?
To calculate the molarity of the acetic acid solution, you can use the concept of stoichiometry in a titration experiment.
First, we need to write a balanced chemical equation for the reaction that takes place between KOH (potassium hydroxide) and acetic acid (CH3COOH). The balanced equation is:
CH3COOH + KOH -> CH3COOK + H2O
From the balanced equation, we can see that one mole of acetic acid reacts with one mole of KOH to produce one mole of water and one mole of potassium acetate. This means that the ratio of acetic acid to KOH is 1:1.
We are given the volume (V1) and molarity (M1) of the KOH solution, as well as the volume (V2) of the acetic acid solution.
V1 = 14.7 mL (volume of KOH solution)
M1 = 0.0550 M (molarity of KOH solution)
V2 = 55.5 mL (volume of acetic acid solution)
Now, we can use the equation:
M1V1 = M2V2
where M2 is the molarity of the acetic acid solution we want to calculate.
Substituting the known values:
(0.0550 M)(14.7 mL) = M2(55.5 mL)
Now, we can solve for M2:
M2 = (0.0550 M)(14.7 mL) / (55.5 mL)
M2 = 0.0145 M
Therefore, the molarity of the acetic acid solution is 0.0145 M.