Calculate the volume of a 0.225 M solution of potassium hydroxide,KOH,required to react with 0.215g of acctic acid,CH,COOH=KCH,COO+H2O.

mols acetic acid (HAc) = grams/molar mass = ?

mols KOH = mols HAc (look at the coefficients in the balanced equation.
M KOH = mols KOH/L KOH. You know mols and M, solve for L and convert to mL if needed.

0.1 ml

To calculate the volume of a solution required for a reaction, we need to use the equation and stoichiometry of the reaction.

The balanced chemical equation for the reaction between potassium hydroxide (KOH) and acetic acid (CH3COOH) is:
CH3COOH + KOH → KCH3COO + H2O

From the balanced equation, we can see that 1 mole of acetic acid reacts with 1 mole of potassium hydroxide.

Step 1: Calculate the number of moles of acetic acid (CH3COOH):
Given mass of acetic acid = 0.215 g
Molar mass of acetic acid (CH3COOH) = 60.05 g/mol

Number of moles of acetic acid = mass / molar mass
Number of moles of acetic acid = 0.215 g / 60.05 g/mol

Step 2: Calculate the volume of the solution (KOH) needed for reaction:
Given molarity of KOH solution = 0.225 M
Number of moles of KOH = Number of moles of acetic acid (from step 1)

We can use the formula: Molarity (M) = Moles / Volume (L) to find the volume of the solution.

Molarity of KOH = Number of moles of KOH / Volume (L)
0.225 M = Number of moles of KOH / Volume (L)

Rearranging the equation:
Volume (L) = Number of moles of KOH / Molarity of KOH
Volume (L) = 0.215 g / 60.05 g/mol / 0.225 mol/L

Step 3: Calculate the volume of the solution:
Volume (L) = 0.215 g / (60.05 g/mol * 0.225 mol/L)

Now, plugging in the numbers and calculating:
Volume (L) = 0.215 g / (13.513 mol/L)
Volume (L) = 0.0159 L

Therefore, the volume of the 0.225 M solution of KOH required to react with 0.215 g of acetic acid is approximately 0.0159 liters or 15.9 mL.

To calculate the volume of the solution required, we need to determine the number of moles of acetic acid, CH3COOH, and then use the balanced chemical equation to find the stoichiometric ratio between acetic acid and potassium hydroxide, KOH.

Step 1: Calculate the number of moles of acetic acid (CH3COOH)
To find the number of moles, we can use the formula:
moles = mass (g) / molar mass (g/mol)

Given that the mass of acetic acid, CH3COOH, is 0.215 g, we need to determine the molar mass of acetic acid, which is calculated by summing the atomic masses of the elements in the formula CH3COOH:
C: 1 atom x 12.01 g/mol = 12.01 g/mol
H: 3 atoms x 1.01 g/mol = 3.03 g/mol
O: 2 atoms x 16.00 g/mol = 32.00 g/mol

Adding these together, we get:
molar mass of CH3COOH = 12.01 g/mol + 3.03 g/mol + 32.00 g/mol = 47.04 g/mol

Now, we can calculate the number of moles:
moles of CH3COOH = 0.215 g / 47.04 g/mol

Step 2: Use the balanced chemical equation to determine the stoichiometric ratio
The balanced chemical equation for the reaction between acetic acid (CH3COOH) and potassium hydroxide (KOH) is:
CH3COOH + KOH → KCH3COO + H2O

The equation tells us that each 1 mole of CH3COOH reacts with 1 mole of KOH. This implies a 1:1 stoichiometric ratio.

Step 3: Calculate the volume of 0.225 M KOH solution required
The concentration of a solution is defined as the number of moles of solute divided by the volume of the solution in liters.

Given that the concentration of the KOH solution is 0.225 M, this means 0.225 moles of KOH are present per liter of solution.

Using the stoichiometric ratio from Step 2, we can determine that the number of moles of KOH required is equal to the number of moles of CH3COOH.

Therefore, the volume of the solution required (in liters) can be calculated using the formula:
volume (L) = moles / concentration (M)
volume (L) = (0.215 g / 47.04 g/mol) / 0.225 M

Finally, calculate the volume:
volume (L) = (0.215 g / 47.04 g/mol) / 0.225 M