Question

DATA

Beverage sample

volume of sample : 79.5 mL Name of beverage : 7up

Solution preparation

Concentration of stock NAOH solution: 0.04 M

Volume of stock NaOH soln. used : 3.2

Mas of KHP calculated to neutralize 15 mL of ~ 0.04 M NaOH : 0.1225

Solution Standardization

Trial 1 : phthalate .127 g

Trial 1: Final volume NaOH 12.6 mL

initial volume NaOH 24.4 ml

Trial 2

.121 g of phthalate

final volume 16.5 mL

Initial volume 24.5 mL

Determination of Citric Acid

Trial 1 : 100 ml

final volume NaOH 9.1

initial volume NaOh 20.8 ml

Trial 2 : 100 ml

final volume NaOH 14.9

Initial volume NaOH 24.6

Determination of beverage sample

trial 1

beverage 36 ml

final volume NaOH 1.6 ml

initial volume NaOH 14.8 ml

Trial 2

35 ml

final volume NaOH 1.4 ml

initial volume 12.3 ml

Analysis:

question 1.

A. Solution standardization

standardized concentration of NaOH solution:

Trial 1: 10.0527 M Trial 2: 20.074 M

average standardized concentration of NaOH solution _ M ?

B. Determinartion of citric acid

Trial 1

moles of citric acid _: 0.000742

moles of citric acid _ moles?: 0.000801

Trial 2

moles of citric acid _ moles?: 0.000615

moles of citric acid _ moles? 0.000700

Please help with the following:

1.       Relationship betwen moles of citric acid and moles of NaOH

Number of moles of NaOH per molecule of citric acid:

Number of moles of citric acid per mole of NaOH

proposed equation for the reaction betwen NaOH and citric acid

Number of acidic hydrogen atoms per molecule of citric acid:

Number of molecules of citric cid per acidic hydrogen atom:

Determination of beverage sample

Moles of NaOH consumed: Trial (1) _ moles and Trial (2) _ moles

Moles of citric acid neutralized: : Trial (1) _ moles and Trial (2) _ moles

Molarity of citric acid in beverage sample : trial (1) _M and trial (2) _M

Average M of beverage : _M

% concentration by mas: (assume density of 1.00 g/mol) _%

Average Concentration:

Answer 1

Balanced reaction between Citric acid and NaOH

Relationship betwen moles of citric acid and moles of NaOH : 1 mol citric acid reacts with 3 moles of NaOH

Number of moles of NaOH per molecule of citric acid : three

Number of moles of citric acid per mole of NaOH : 1/3 moles

proposed equation for the reaction betwen NaOH and citric acid : shown above

Number of acidic hydrogen atoms per molecule of citric acid: three

Number of molecules of citric acid per acidic hydrogen atom: 1/3

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Determination of beverage sample

Moles of NaOH consumed: Molarity * volume in L

Trial 1 : 0.04 M * 0.0132 L = 5.28*10^-4 moles

Trial 2 : 0.04 * 0.0109 L = 4.36 *10^-4 moles

Moles of citric acid neutralized: : Trial (1) 5.28*10^-4 moles/3 =1.76*10^-4 mol

                                                       Trial (2) 4.36 *10^-4 moles /3= 1.45 *10^-4

Molarity of citric acid in beverage sample : trial (1) _M = 1.76*10^-4 mol *1000mL/36mL = 0.0048

                                                             trial (2) M= 1.45 *10^-4 *1000/35ml = 0.0041M

Average M of beverage : _M = 0.00445 M

% concentration by mas: (assume density of 1.00 g/mol) _%

Trial 1 :

mass in 36 mL = 1.76*10^-4 mol *192.124g/mol = 0.034g

% concentration = 0.034 *100mL/35mL = 0.097%

Trial 2 :

mass in 36mL = 1.45 *10^-4 *192.124 f/mol = 0.0278 g

% concentration = 0.0278g *100mL/36mL = 0.077 %

Average Concentration: