25.00 mL of an acetic acid solution requires 23.86mL of a 0.1550M potassium hydroxide solution to reach the end point. What is the molarity of the acetic acid solution? (Hint: write the chemical reaction, and remember stoichiometry)
Write and balance the equation.
moles KOH = M x L
Use the equation to convert moles KOH to moles acetic acid.
Macetic acid = moles/L
C2H4O2 + KOH-->
I'm not quite sure how you balance the equation?
The empirical formula you have written tells you nothing about the molecule (at least not much except for the molar mass). If you write it as CH3COOH OR as HC2H3O2, it is easier to see that the H on the right most end of the CH3COOH is the H that ionizes or it's the left most H on HC2H3O2 so the ionization is
CH3COOH ==> H^+ + CH3COO^- or
HC2H3O2 ==> H^+ + C2H3O2^-
the C2H3O2 is the acetate ion (CH3COO^-). Frankly, I reduce it even further because I get tired of typing all that "other" stuff. I just call it HAc where the H^+ stands for hydrogen and the Ac^- stands for the acetate ion.
The neutralization reaction is
HAc + KOH ==> KAc + H2O.
To find the molarity of the acetic acid solution, we need to use the stoichiometry of the chemical reaction between acetic acid and potassium hydroxide. The balanced chemical equation for this reaction is:
CH3COOH + KOH => CH3COOK + H2O
From the balanced equation, we can see that the moles of acetic acid (CH3COOH) are equal to the moles of potassium hydroxide (KOH) required to react:
1 mol CH3COOH = 1 mol KOH
Given that 23.86 mL of a 0.1550 M KOH solution is required to reach the endpoint, we can calculate the moles of KOH used:
moles of KOH = volume (L) x concentration (mol/L)
= 0.02386 L x 0.1550 mol/L
= 0.00369 mol KOH
Since the moles of acetic acid and KOH are equal, we now know the moles of acetic acid used in the reaction:
moles of CH3COOH = 0.00369 mol
Now, we need to calculate the volume of the acetic acid solution used. Given that the volume is 25.00 mL, we convert it to liters:
volume of CH3COOH = 25.00 mL x (1 L /1000 mL)
= 0.02500 L
The molarity of the acetic acid solution is then:
molarity of CH3COOH = moles of CH3COOH / volume of CH3COOH
= 0.00369 mol / 0.02500 L
= 0.1476 M
Therefore, the molarity of the acetic acid solution is 0.1476 M.