please help me with #2 I don\'t understand this problem in an experiment to dete

Question

please help me with #2 I don't understand this problem

in an experiment to determine the molecular weight and the ioeizarion ascotie acid (vitamin s for milliliters of solani 13717 grams of the water to make 50.00 The n. The entire solution was ed with a 0.22 molar monitored throughout the The equivalence point was reached when been added. the this ascortic acid acts as a monoprotic acid can be represented as HA a) From the information above calculate the molecular weight of ascortic acid. b) when 20.00 milliliters of NaoH had been added during the titration, the pH of the solution was 4.23. Calculate the acid nization constant for ascorbic acid c) Calculate the K for the ascorbate ion, A dy Calculate the pH of the solution at the equivalence point of the titration

Explanation / Answer

From the information above, calculate the molecular weight of ascorbic acid

HA + NaOH --> NaA + H2O                              Neutralization Reaction of monoprotic acid

1 mol HA = 1 mol NaOH                                          Coefficient Equality
1.3717 g HA = 35.23 ml NaOH                                Set mass and volume equal
0.2211 mol NaOH = 1000 ml NaOH                        Molarity Equality
   

               Goal =  g HA/ mol HA

(1.3717 g HA    / 35.23 ml NaOH   )*( 1 mol NaOH/ 1 mol HA)*( 1000 ml NaOH/ 0.2211 mol NaOH)

= 176.1 g/mol----------------ANSWER

===========================================================

When 20.0 ml of NaOH had been added during titration, the pH of the solution was 4.23. Calculate the acid ionization constant of ascorbic acid.

Although this is not exactly at the midpoint of the titration, the pH changes very little in this range.

pH = pKa = 4.23           

        pH = pKa at the midpoint of a titration

pKa = - log Ka  => Ka = antilog ( - pKa )

= antilog ( ( -4.23) = 5.9 x 10 -53) =

===================================

Calculate the equilibrium constant for the reaction of the ascorbate ion, A-, with water.

K b A- x   K a HA = 1.00 x 10 ^-14

A -1 + H2O -----> HA + OH -1 Basic Salt Reaction

K b A= 1.00 x 10^ -14   / K a HA

= 1.00 x 10 ^-14     / 5.9 x 10^ -5  

= 1.7 x 10^ -10

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Calculate the pH of the solution at the equivalence point of the titration.
    

At the equivalence point, only the salt, NA, is present. It is dissociated into Na + and A -1.
   [A -1]= 0.2211 M x 25.23 ml / 75.23 ml= = 0.07415 M           Dilution Equation

A -1

H2O

HA

OH -1  

i    

0.07415   

0   

0

/

X

X

X

e

0.07415 - X

X

X

Kb = 1.7 x 10 -10 = HA][OH -1]   / [A -1 ] = X . X/ 0.07415

X = ( 0.07415 x 1.7 x 10 ^-10 ) 1/2 = 3.6 x 10^ -6 M = [OH -1 ]

pOH = - log [OH -1] = - log 3.6 x 10 -6 M = 5.44

14.00 = pH + pOH  => pH = 14.00 - 5.44 = 8.56

pH = 14.00 - 5.44 = 8.56

Calculate the pH of the solution at the equivalence point of the titration.
    

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