Titration of Acetic Acid Solution using Sodium Hydroxide LAB Please help with ca

ID: 694996 • Letter: T

Question

Titration of Acetic Acid Solution using Sodium Hydroxide LAB

Please help with calculation (and show the work) thank you.

The sodium hydroxide solution was made by dissolving 25.9865g of NaOH in 2.00L of water.

Calculate the molarity (M) of NaOH ___________ ?

Molarity of Acetic Acid in Vinegar

Volume of Acetic Acid (vinegar) 10.00mL

Initial volume of NaOH 0

Final volume of NaOH 29.00mL

Calculations:

Volume of NaOH used _______?

Moles of NaOH _______?

Moles of Acetic Acid _______?

Molarity of Acetic Acid _______?

Average Molarity of Acetic Acid _______?

Important formulas

Molarity = Moles/L

Mass % = mass of KHP / mass of KHP solution x 100%

At equivalent point: moles NaOH = moles acetic

NaOH + CH3COOH ------> CH3COONa + H2O

Molarity of NaOH = mass of the solute/molar mass of thr solute * 1/Volume of the solution in L

Molarity of NaOH = 25.9865/40 * 1/2

Molarity of NaOH = 0.3248 mol/L

number of moles of NaOH = 0.3148 * 29 * 10^-3 = 9.1292 * 10^-3 moles

1 mole of NaoH requires 1 mole of CH3COOH according to the equation,

so number of moles of Acetic acid = 9.1292 * 10^-3 moles

molarity of acetic acid = number of moles * 1000/volume of solution in mL

Molarity of Acetic acid = 9.1292 * 10^-3 * 1000/10

Molarity = 0.9129 mol/L

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