How to calculate the pH? & which would be the best buffer

Chemistry
Tutor: None Selected Time limit: 1 Day

1. Acetylsalic acid is a week acid (HC9H7O4 Pka=3.48) A 25 mL solution of 0.15 M aspirin is titrated with 0.10 M NaOH. calculate the pH of the solution after 15 mL of NaOH is added and the volume of NaOH needed to reach the equivalence point of the titration.

2.which of the following acid-base solutions would be the most effective buffer when dissolved in water?

A. .50M CH3COOH & .10M CH3COONa

B. .30 M CH3COOH & .30 M CH3COONa

C. .30M NaCl & .30 M HCl

D. .20M NH3 & .40M HCl

E. .40M HBr & .20M NaOH

May 12th, 2015

DATA:

Volume of aspirin =25mL

Concentration of aspirin= 0.15M

Volume of NaOH= 15mL

Concentration of NaOH= 0.1M

pH=?

SOLUTION:

Moles of NaOH= M* volume in L

  = (0.1*15)/1000

  = 0.0015 mol

Moles of acetylsalicylic acid= M * volume in L

  = (0.15*25)/1000

  = 0.00375mol

1NaOH = 1 acetylsalicylic acid

0.0015 mol= 0.0015 mol of acetylsalicylic acid

Moles remaining = total moles – reacted moles

  = 0.00375 – 0.0015

  = 0.0022 moles of acetylsalicylic acid

pH= -log [H+]

pH= -log 0.0022

  = 2.6 

part B:

0.30M acetic acid and 0.30M sodium acetate will be the best buffer because both the acid nad its conjugate base are in same amount and it is a criteria for selecting buffer.

May 12th, 2015

i wanna correct the last part of my calculation. it'd be done like this:

1NaOH = 1 acetylsalicylic acid

0.0015 mol= 0.0015 mol of acetylsalicylic acid

Moles remaining = total moles – reacted moles

  = 0.00375 – 0.0015

  = 0.0022 moles of acetylsalicylic acid

M of acetylsalicylic acid= mol/volume in L

  = (0.0022*1000)/25

  = 0.088 M

pKa= -log Ka

3.48 = -logka

Ka=antilog – 3.48

Ka= 3.311*10-4

Ka = [C9H7O4-] [H+]/[HC9H7O4]

Suppose [C9H7O4-] =[H+]= x

  3.311*10-4= x2/ 0.088

X2= 2.913*10-5

X=  5.39*10-3M

pH=-log [H+]

  = -log 2.913*10-5

  = 2.2 

i hope it helps you understand. if there is any confusion you can ask me


May 12th, 2015

so at the very end why did you do -log (2.913 *10-5) instead of the -log(5.39*10-3)

May 12th, 2015

oh....it's a typing mistake....you are right, it'd be -log 5.39*10^-3 and the result will be 2.2

May 12th, 2015

and did you come up with how to find the NaOh needed to reach the equivalence point because im really confused about that part

May 12th, 2015

  Moles of acetylsalicylic acid= M * volume in L

  = (0.15*25)/1000

  = 0.00375mol

1 acetylsalicylic acid = 1 NaOH

0.00375 mol = 0.00375 mol of NaOH at equivalence point

M of NaOH = mol of NaOH/ Volume of NaOH in L

Volume of NaOH in L = mol of NaOH/ M of NaOH

  = 0.00375/ 0.1

  = 0.0375 L

  = 37.5 mL 


May 12th, 2015

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