What is the molarity of a solution of acetic acid if 35.00mL is titrated to the end point with 68.20ml of 0.750M KOH? What is the percentage by mass of acetic acid solution? The density of acetic acid is 1.06 g/ml

acetic acid is HAc

HAc + KOH ==> KAc + H2O
mols KOH = M x L
Look at the coefficients. mols HAc = mols KOH
Then M HAc = mols HAc/L HAc

To find the molarity of the acetic acid solution, you can use the formula:

M1V1 = M2V2

Where:
M1 = molarity of the KOH solution
V1 = volume of the KOH solution (in liters)
M2 = molarity of the acetic acid solution (what we want to find)
V2 = volume of the acetic acid solution (in liters)

In this case, M1 = 0.750 M, V1 = 68.20 mL (convert to liters by dividing by 1000), V2 = 35.00 mL (convert to liters by dividing by 1000).

So the equation becomes:

0.750 M x (68.20 mL / 1000 mL) = M2 x (35.00 mL / 1000 mL)

Now, solve for M2:

M2 = (0.750 M x 68.20 mL) / 35.00 mL

Now, let's calculate the value of M2:

M2 = 1.459 M

Therefore, the molarity of the acetic acid solution is 1.459 M.

To find the percentage by mass of acetic acid in the solution, we need to first calculate the mass of acetic acid present in the solution. The formula for calculating mass is:

Mass = Volume x Density

Given that the volume of the solution is 35.00 mL and the density of acetic acid is 1.06 g/mL, we can find the mass:

Mass = 35.00 mL x 1.06 g/mL
Mass = 37.10 g

Next, to find the percentage by mass, we need to divide the mass of the acetic acid by the total mass of the solution, and then multiply by 100:

Percentage by mass = (Mass of acetic acid / Total mass of solution) x 100

Given that the mass of the acetic acid is 37.10 g (as calculated before), and the total mass of the solution is 35.00 mL (which is equivalent to 35.00 g, as the density of acetic acid is 1.06 g/mL), we can calculate the percentage by mass:

Percentage by mass = (37.10 g / 35.00 g) x 100
Percentage by mass = 105.71%

Therefore, the percentage by mass of acetic acid in the solution is 105.71%.