For many weak acid or weak base calculations, you can use a simplifying assumption to avoid solving quadratic?

2 Answers

• The accepted rule is that you do not have to use the more complicated quadratic equation if the % dissociation is less than 5%.

  Let us work out a general equation to calculate the [H+] in each case

  Ka = [H+]² / [acid]

  [H+]² = Ka * [acid]

  [H+] = √( Ka * [acid])

  Now apply this to each case and then calculate % dissociation:

  a) [HA]=0.01M Ka=1x10^ -4:

  [H+] = √ (110^-40.01)

  [H+] = √ ( 1*10^-6)

  [H+] = 1*10^-3

  % dissociation : (110^-3) / 0.01100 = 10% - you must use quadratic

  b) [HA]=0.01M Ka=1x10^ -5:

  [H+] = √(Ka * [acid])

  [H+] = √ ( 110^-5 0.01)

  [H+] = √ 1*10^-7

  [H+] = 3.16*10^-4

  % dissociation = (3.1610^-4)/ 0.01100 = 3.16% = Do not use quadratic

  c. [HA]=0.1M Ka=1x10^ -3:

  [H+] = √ ( 110^-3) 0.1

  [H+] = √ 1*10^-4

  [H+] = 1*10^-2

  % dissociation = (110^-2) / 0.1 100 = 10% - you must use quadratic

  d.[HA]=1M Ka=1x10^ -3:

  [H+] = √( 110^-3 1)

  [H+] = √ 1*10^-3

  [H+] = 0.0316

  % dissociation = 0.0316/1*100 = 3.16% = no need to use quadratic

  e. [HA]=0.001M Ka=1x10^ -5

  [H+] =√ (110^-5 0.001)

  [H+] = √ 1*10^-8

  [H+] = 1*10^-4

  % dissociation = (110^-4) / 0.001100 = 10% You must use quadratic.

  Assumption valid - no need to use quadratic = B and D

  Assumption mnot valid - you must use quadratic = A, C ,E

• Calculations of volumetric nature contain the comparable formulae whether the acids or bases in touch are vulnerable or sturdy.i.e. N1V1 = N2V2. The pH exchange close to the top ingredient is slow and not sharp in spite of the incontrovertible fact that in case you would be able to desire to calculate pH on the neutralization ingredient then you certainly would use the formulation for looking pH of a salt of vulnerable acid and vulnerable base given under: pH = 7 + a million/2 pKa - a million/2pKb