Question

DATA SHEE Calculate the mass percent acetic acid present in a sample of balsamic vinegar if 20.72 mL of 0.125 M sodium hydroxide was needed to neutralize the acetic acid present in 3.281 g of the vinegar 3. 4. Find the mililiters of 0.220 M KOH needed to neutralize 85.0 mL of 0.816 M H,PO, g one chain of conversion Balance the chemical equation and show your work usin factors in your calculation.

Answer 1

1. % acetic acid = (Weight of acetic acid in sample / Total weight of sample) * 100

CH3COOH + NaOH CH3COONa + H2O

% acetic acid = (Weight of acetic acid in sample/Total weight of sample) * 100 CH3COOH + NaOH rightarrow CH_3COONa + H_2O Hence number of moles of CH_3COOH = number of moles of NaOH Given, V = 20.72 mL = 0.02072 L M = 0.125 M Number of moles of NaOH = 0.125 mol/L times 0.02072 L = 0.00259 mol Hence number of moles of CH_3COOH = 0.00259 mol Molar mass of CH_3COOH = 60.052 g/mol Hence mass of CH3COOH = Molar mass times Number of moles = 60.052 g/mol times 0.00259 mol = 0.155 g Weight of sample = 3.281 g Hence % mass = (0.155/3.281) * 100 = 4.72 % H_3PO4 + 3 KOH rightarrow K_3PO_4 + 3 H_2O Here we will use the formulae M_1V_1 = M_2V_2 where M_1 = 0.220 M V_1 = Unknown M_2 = 0.816 M V_2 = 85 mL = 0.085 L Substituting the values in the above equation, we get V_1 times 0.220 = 0.816 times 0.085 V_1 = 0.315 L = 315 mL Hence 315 mL of KOH will be required.

Hence number of moles of CH3COOH = number of moles of NaOH

Given, V = 20.72 mL = 0.02072 L

M = 0.125 M

Number of moles of NaOH = 0.125 mol/L x 0.02072 L

= 0.00259 mol

Hence number of moles of CH3COOH = 0.00259 mol

Molar mass of CH3COOH = 60.052 g/mol

Hence mass of CH3COOH = Molar mass x Number of moles

= 60.052 g/mol x 0.00259 mol

= 0.155 g

Weight of sample = 3.281 g

Hence % mass = (0.155 / 3.281) * 100 = 4.72 %

2. H3PO4 + 3 KOH
% acetic acid = (Weight of acetic acid in sample/Total weight of sample) * 100 CH3COOH + NaOH rightarrow CH_3COONa + H_2O Hence number of moles of CH_3COOH = number of moles of NaOH Given, V = 20.72 mL = 0.02072 L M = 0.125 M Number of moles of NaOH = 0.125 mol/L times 0.02072 L = 0.00259 mol Hence number of moles of CH_3COOH = 0.00259 mol Molar mass of CH_3COOH = 60.052 g/mol Hence mass of CH3COOH = Molar mass times Number of moles = 60.052 g/mol times 0.00259 mol = 0.155 g Weight of sample = 3.281 g Hence % mass = (0.155/3.281) * 100 = 4.72 % H_3PO4 + 3 KOH rightarrow K_3PO_4 + 3 H_2O Here we will use the formulae M_1V_1 = M_2V_2 where M_1 = 0.220 M V_1 = Unknown M_2 = 0.816 M V_2 = 85 mL = 0.085 L Substituting the values in the above equation, we get V_1 times 0.220 = 0.816 times 0.085 V_1 = 0.315 L = 315 mL Hence 315 mL of KOH will be required.
K3PO4 + 3 H2O

Here we will use the formulae M1V1 = M2V2

where M1 = 0.220 M

V1 = Unknown

M2 = 0.816 M

V2 = 85 mL = 0.085 L

Substituting the values in the above equation, we get

V1 x 0.220 = 0.816 x 0.085

V1 = 0.315 L = 315 mL

Hence 315 mL of KOH will be required.