A 15.5 mL sample of 0.215 M KOH solution requires 21.2 mL of acetic acid in a titration experiment. Calculate the molarity of the acetic acid solution. Please show all steps, I don't understand this.

CH3COOH + KOH ==> CH3COOK + H2O

moles KOH = M x L = ??
Convert moles KOH to moles CH3COOH. Since the coefficients are one in the equation, moles KOH = moles CH3COOH.
Then M acid = moles acid/L acid.

the concentration of HCl, a strong acid is 0.500 M. if 20.0 ml HCl is needed to titrate 40.0 ml of NaOH, what is the concentration of NaOH

I don't understand how to do this. Can I see a step by step way of how to do this please

To calculate 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
V2 = volume of the acetic acid solution (in liters)

Given:
M1 = 0.215 M (molarity of the KOH solution)
V1 = 15.5 mL (volume of the KOH solution)
V1 = 15.5 mL / 1000 mL/L = 0.0155 L (volume of the KOH solution in liters)
V2 = 21.2 mL (volume of the acetic acid solution)
V2 = 21.2 mL / 1000 mL/L = 0.0212 L (volume of the acetic acid solution in liters)

Let's plug in these values into the formula:

0.215 M * 0.0155 L = M2 * 0.0212 L

Solving for M2:

0.0033325 = M2 * 0.0212

Divide both sides of the equation by 0.0212:

M2 = 0.0033325 / 0.0212

M2 = 0.157 M

Therefore, the molarity of the acetic acid solution is 0.157 M.

To calculate the molarity of the acetic acid solution, you need to use the concept of stoichiometry and the equation of the reaction between KOH and acetic acid. Here are the steps:

Step 1: Write the balanced chemical equation for the reaction between KOH and acetic acid:
KOH + CH3COOH -> KCH3COO + H2O

Step 2: Determine the stoichiometry of the reaction. From the balanced equation, you can see that one mole of KOH reacts with one mole of acetic acid.

Step 3: Calculate the number of moles of KOH used. Use the formula:
moles of KOH = Molarity of KOH solution × Volume of KOH solution (in liters)

Given that the volume of the KOH solution is 15.5 mL = 0.0155 L and the molarity of the KOH solution is 0.215 M, substitute the values into the formula:
moles of KOH = 0.215 M × 0.0155 L = 0.0033325 moles

Step 4: Determine the number of moles of acetic acid based on the stoichiometry of the reaction. Since the stoichiometry is 1:1, the number of moles of acetic acid is also 0.0033325 moles.

Step 5: Calculate the molarity of the acetic acid solution. The formula for molarity is:
Molarity = Moles of solute / Volume of solution (in liters)

Given that the volume of the acetic acid solution is 21.2 mL = 0.0212 L, substitute the values into the formula:
Molarity = 0.0033325 moles / 0.0212 L = 0.157 M

Therefore, the molarity of the acetic acid solution is 0.157 M.