Help me please! Joe Smith performed a series of acid-base reactions to determine

Question

Help me please!

Joe Smith performed a series of acid-base reactions to determine the percentage (by mass) of the acid in his vinegar, 25.49 mL of NaOH were required to neutralize 0.5208 g of KHP (molar mass = 204.22 g/mol) dissolved in water. This titration to standardize the NaOH. The vinegar solution for titration was prepared in the following manner: A volume of 25.0 mL of vinegar (density = 1.00 g/mL) was diluted 250.0 mL integral a volumetric flask, and 25.0 mL of this diluted solution required 22.62 mL of the above standardized NaOH to reach the phenolphthalein endpoint. Show your calculations in detail for each question below. What was the molarity of the NaOH solution used? What is the percentage by mass of the acid in the original vinegar sample?

Explanation / Answer

(a) Write down the balanced chemical reaction between KHP and NaOH.

KHP + NaOH -------> NaKP + H2O

As per the balanced stoichiometric reaction,

1 mole KHP = 1 mole NaOH

Calculate the moles of KHP used = (mass of KHP/molar mass of KHP) = (0.5208 g)/(204.22 g/mol) = 0.002550 mole.

Therefore, moles of NaOH required for the titration = (0.002550 mole KHP)*(1 mole NaOH/1 mole KHP) = 0.002550 mole.

Molarity of NaOH used = (moles of NaOH required)/(volume of NaOH in L) = (0.002550 mole)/[(25.49 mL)*(1 L/1000 mL)] = 0.10003 mol/L 0.100 M (ans).

(b) Volume of stock vinegar solution taken = 25.0 mL.

Density of stock vinegar solution taken = 1.00 g/mL.

Therefore, mass of stock vinegar solution taken = (25.0 mL)*(1.00 g/mL) = 25.00 g.

25.0 g of the stock vinegar solution was dissolved in 250.0 mL volumetric flask and 25.0 mL of the diluted solution required 22.62 mL of the standardized NaOH solution.

Write down the balanced chemical equation for the reaction:

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

As per the stoichiometric reaction,

1 mole NaOH = 1 mole acetic acid (main component of vinegar which is titrated with NaOH).

Use the dilution law to calculate the concentration of acetic acid in vinegar:

M1*V1 = M2*V2 where M1 = molarity of NaOH solution, V1 = volume of NaOH solution; M2 = molarity of diluted acetic acid solution in vinegar and V2 = volume of vinegar solution required.

Therefore,

(0.100 M)*(22.62 mL) = M2*(25.0 mL)

===> M2 = (0.100 M)*(22.62 mL)/(25.0 mL) = 0.09048 M.

Next calculate the moles of vinegar in the diluted 25 mL solution used for titration.

Moles of vinegar = (volume of solution in L)*(molar concentration of vinegar) = (25 mL)*(1 L/1000 mL)*(0.09048 mol/L) = 0.002262 mole.

Calculate the moles of vinegar in the diluted 250.0 mL solution = (0.002262 mole)*(250 mL/25 mL) = 0.02262 mole.

Use the atomic mass of the elements present in acetic acid to compute the molar mass of acetic acid (C2H4O2) = (12*2 + 1*4 + 16*2) g/mol = 60 g/mol.

Calculate the mass of acetic acid present in vinegar as (moles of acetic acid)*(molar mass of acetic acid) = (0.02262 mole)*(60 g/mol) = 1.3572 g.

Calculate the percentage of acetic acid in the vinegar sample = (mass of acetic acid)/(mass of vinegar)*100 = (1.3572 g)/(25.0 g)*100 = 5.4288% 5.43% (ans).

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