What is the molarity of an Acetic Acid Solution that requires 40.0 ml of 0.100M NaOH to titrate a 20.0ml sample to the end point?

(Balanced equation & solution please! Thanks in advance)

Refer to your question above for Ca(OH)2 and HCl.

To find the molarity of the Acetic Acid Solution, we can use the concept of stoichiometry and the balanced chemical equation between acetic acid (CH3COOH) and sodium hydroxide (NaOH).

Here is the balanced chemical equation for the reaction:

CH3COOH + NaOH -> CH3COONa + H2O

From the balanced equation, we can determine the stoichiometric ratio between CH3COOH and NaOH, which is 1:1.

Given that 40.0 mL of 0.100 M NaOH is required to titrate a 20.0 mL sample, we can use the equation:

(moles of CH3COOH) x (volume of Acetic Acid Solution) x (molarity of NaOH) = (moles of NaOH) x (volume of NaOH)

Let's plug in the values:

(x) x (20.0 mL) x (0.100 M) = (0.100 M) x (40.0 mL)

The volume units need to be the same, so let's convert mL to L:

(x) x (0.020 L) x (0.100 M) = (0.100 M) x (0.040 L)

Simplifying the equation:

0.002x = 0.004

Now, solve for x:

x = 0.004 / 0.002

x = 2

Therefore, the molarity of the Acetic Acid Solution is 2.000 M.