DATA Beverage sample volume of sample : 79.5 mL Name of beverage : 7up Solution

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 _ M ? Trial 2 _ M ?

average standardized concentration of NaOH solution _ M ?

B. Determinartion of citric acid

Trial 1

moles of citric acid _ moles?

moles of citric acid _ moles?

Trial 2

moles of citric acid _ moles?

moles of citric acid _ moles?

C. relationshipbetwen 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:

Explanation / Answer

A) 1) molar mass of KHP(potassium hydrogen pthalate)=204.22 g/mol

So moles of KHP=mass /molar mass=0.127g/204.22 g/mol=0.000622 moles

volume of NaOH used up=final volume-initial volume=24.4-12.6=11.8 ml

So moles of NaOH neutralized=moles of KHP=0.000622 moles

Std molarity of NaOH=0.000622 moles/11.8ml *(1000 ml/L) =0.0527 moll/L

2) volume of NaOH=final volume-initial volume=24.5-16.5=8 ml

moles of KHP=mass /molar mass=0.121 g/204.22 g/mol=0.000592 moles

As moles of NaOH neutralized=moles of KHP=0.000599 moles

so,Std molarity of NaOH=0.000599 moles/8ml *(1000 ml/L) =0.0741 moll/L

standardized concentration of NaOH solution:

Trial 1 0.0527 M

Trial 2 0.0741M

average standardized concentration of NaOH solution _=0.0527+0.0741/2=0.0634M

B)

Determination of Citric Acid

Trial 1 : volume of Citric acid=V1=100 ml

molarity of citric acid=M1

volume of NaOH=20.8-9.1=V2=11.7 ml

std molarity of NaOH=M2=0.0634M

M1*V1=M2*V2=moles reacted for both =0.0634 mol/L*11.7ml* (1L/1000ml)=0.000742 moles

moles of citric acid _ moles?=0.000742 moles

trial 2

volume of NaOH=24.6-14.9=9.7 ml

M1*V1=M2*V2=moles reacted for both =0.0634 mol/L*9.7ml* (1L/1000ml)=0.000615 moles

moles of citric acid _ moles?=0.000615 moles

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