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pH and Buffers (Example questions (Weak acid (Ka from pKa (Calculate the…

- - - - If HCl has a pH of 2 what is its concentration?
      - 2 = - log[H+]
      - -2 = log[H+]
      - [H+] = 10^-2 = 0.01
    - - If HCl has a concentration of 0.5M what is it's pH?
      - pH = -log[H+]
      - pH = -log0.5
      - pH = - - 0.30
      - pH = 0.3
  - - - If CH3COOH has a Ka of 1.7 x 10^-5 and a concentration of 0.5 M what is it's pH?
      - Ka = [H+][A-]/[HA] as [H+]=[A-] this becomes Ka = [H+]^2/[HA]
      - 1.7x10^-5 = [H+]^2/0.5
      - 8.5x10^-6 = [H+]^2
      - [H+] = 2.9x10^-3
      - -log[H+] = -log[2.9x10^-3] = 2.54
    - - Calculate the pH of 2M HA, pKa of 4.8
      - pKa = -logKa
      - Ka = 10^-4.8 = 1.58x10^-5
      - 1.58x10^-5 = [H+]^2/[HA]
      - [H+] = 5.6x10^-3
      - -log[H+] = 2.25
      - 1.58x10^-5 = [H+]^2/[HA]
    - - If the concentration of HA is originally 0.8M and it dissociates to form 0.004M of H+ and A- each, what is its Ka
      - Ka = [H+][A-]/[HA]
      - Ka = 0.004*0.004/(0.8-0.004) = 2x10^-5
  - - - Find the pH of 1M NaOH
      - 14 -pOH = pH
      - pOH = -log[OH-] = -log1 = 1
      - 14-1 = 13
  - - - A buffer consists of 0.4M ethanoic acid and 0.2M sodium ethanoate, calculate the pH of the buffer. (Ka of ethanoic acid is 1.7 x 10^-5)
      - Ka = [H+][A-]/[HA]
      - Sodium ethanoate fully dissociates so 0.2M
      - 1.7x10^-5 = [H+]*0.2/0.4
      - 1.7x10^-5 x 2 = [H+] = 3.4x10^-3
      - Ethanoic acid is almost undissociated so ≈ 0.4M
      - -log[H+] = 4.47
    - - Calculate the pH of the buffer formed when 0.5dm3 of 0.2M NaOH is added to 0.5dm^3 of 0.6M HA. Ka = 6.25x10^-5
      - Moles HA = 0.6*0.5 = 0.3
      - Moles NaOH = 0.2*0.5 = 0.1
      - HA + NaOH -> H2O + NaA Moles before = 0.3 + 0.1 -> + 0 Moles after = 0.2 + 0 -> + 0.1
      - This leaves a 1dm3 solution of 0.2 moles of HA
      - [HA] = 0.2 [A-] = 0.1
      - Ka = [H+][A-]/[HA]
      - 6.25x10^-5 = [H+]*(0.1/0.2)
      - 1.25x10^-4 = [H+]
      - -log[H+] = 3.9
  - - - Ka for propanoic acid = 1.35x10^-5
      - a) Calculate the pH of 0.169M solution of propanoic acid
      - [H+] = 1.55 x10^-5
      - biii) Ka = [H+][A-]/[HA] 1.35x10^-5 = [H+]x0.205/0.235
      - Moles of propanoic acid - 0.235 Moles of propanoate ions - 0.205
      - bii) Moles before reaction 0.015 + 0.250 -> 0.190 +
        Moles after reaction 0 + 0.235 -> 0.205 +
      - b) i) NaOH + CH3CH2COOH -> NaCH3CH2COO + H2O
      - b) a buffer solution contains 0.250 mol of propanoic acid and 0.190 mol of sodium propanoate in 1dm3 of solution. A 0.015 mol sample of sodium hydroxide is added to this buffer solution i) Write an equation for the reaction of propanoic acid with sodium hydroxide
        ii) Calculate the number of moles of propanoic acid and propanoate ions present in the buffer solution after the addition of the sodium hydroxide
        iii) Hence calculate the pH of the buffer solution after the addition of the sodium hydroxide
      - -log[H+] = 2.82
      - [H+] = 1.5x10^-3
      - 1.35x10^-5 x 0.169 = [H+]^2 = 2.3 x10^-6
      - 1.35x10^-5 = [H+]^2/0.169
      - Ka = [H+][A-]/[HA]
      - pH = -log[H+] = 4.81
    - - a) The ionic product of water has the symbol Kw i) write an expression for the ionic product of water ii) at 42C the value of Kw is 3.46x10^-14 Calculate the pH of water at this temperature iii) At 75C a 0.0470M solution of sodium hydroxide has a pH of 11.36, calculate Kw at this temperature
      - ai) H2O + H2O -> H3O+ + OH-
      - aii) [H+}^2 =3.46x10^-14
      - [H+] = 1.86x10^-7
      - -log[H+] = 6.73
      - [H+]^2 = 9.968x10^-6
      - biii) 1.78x10^-4 = [H+]^2/0.056
      - bii) Ka = [H+][CHOO-]/[HCOOH]
      - bi) HCOOH -> HCOO- + H+
      - b) Methanoic acid (HCOOH) dissociates slightly in aqueous solution. i) write an equation for this dissociation. ii) Write an expression for the equilibrium constant Ka for methanoic acid. iii) The value of Ka for methanoic acid is 1.78x10^-4 at 25C Calculate the pH of a 0.056 M solution of methanoic acid. iv) The dissociation of methanoic acid is in aqueous solution is endothermic, deduce whether the the pH of a solution of methanoic acid will increase, decrease or stay the same if the solution is heated.
      - Ka = [H+][A-]/[HA] so [H+]*0.47 = 2.05x10^-12 = Kw
      - aiii) 10^-11.36 = [H+] = 4.37x10^-12
      - [H+] = 3.16x10^-3
      - -log[h+] = pH = 2.50
      - The pH will increase because the dissociation reaction is endothermic meaning that when more heat is added the equilibrium will shift to oppose the change and the dissociation reaction will occur and thus more H+ ions will be formed and thus the pH will increase.
      - c) The value of Ka for methanoic acid is 1.78x10^-4 at 25C. A buffer solution is prepared containing 2.35x10^-2 moles of methanoic acid and 1.84x10^-2 moles of sodium methanoate in 1dm3 of solution. i) Calculate the pH of this buffer solution at 25C ii) A 5cm3 sample of 0.1M Hydrochloric acid is added to the solution. Calculate the pH of the buffer solution after this addition
      - [H+] = 2.73x10^-4
      - ci) 1.78x10^-4 = [H+]*1.84x10^-2/2.35x10^-2
      - -log[H+] = 3.64
      - cii) 0.1*5/1000 = 5x10^-4 moles of HCl
      - Equation, HCl + CHOO- -> CHOOH + Cl- moles before 0.0005 + 1.84x10^-2 -> 2.35x10^-2 + moles after 0 +0.0179 -> 0.024
      - Ka = [H+][CHOO-]/[HCOOH]
      - 1.78x10^-4 = [H+] * 0.0179/0.024
      - [H+] = 2.39x10^-4
      - -log[H+] = 3.62
    - - a) i) The dissociation of water can be represented by the following reaction H2O -><- H+ + OH- i) write an expression for the ionic product of water, ii) The pH of pure water at 50C is 6.63 Calculate the concentration in M of the H+ ions in this water, iii) Deduce the concentration of OH- ions in this water iv) Calculate the value of Kw at this temperature
      - ai) Kw = [H+][OH-]
      - aii) 10^-6.63 = [H+] = 2.34x10^-7 M
      - aiii) [H+] = [OH-] = 2.34x10^-7 M
      - aiv) (2.34x10^-7)^2 = 5.50x10^-14
      - b) At 25C the value of Kw is 1x10^-14, Calculate the pH of a 0.136M solution of KOH at 25C
      - b) 1x10^-14 = [H+]x0.136
      - [H+] = 7.35x10^-14
      - -log[H+] = 13.13
- - - - So HA -> <- H+ + A- when OH- is added this becomes HA + OH- -> H2O + A- which removes OH- and pH remains almost the same
    - - So HA -> <- H+ + A- when H+ is added it reacts with the extra A- that will have been added to the solution to make the reaction go backwards and the pH does not change.
- - - - e.g. Methyl orange, Phenolphthalein
  - - - e.g. Methyl orange
  - - - e.g. Phenolphthalein