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Acid Strength

Acid Strength. Binary acids. General formula HX ( aq ) , where X= Cl , Br, I and F Two factors for acid strength trend: Across a period, electronegativity increases; the acid strength increases as EN increases

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Acid Strength

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  1. Acid Strength

  2. Binary acids General formula HX(aq), where X=Cl, Br, I and F • Two factors for acid strength trend: • Across a period, electronegativity increases; the acid strength increases as EN increases • WHY? The more electronegative the atom , the more it draws electrons away from the hydrogen atom – making it relatively positive • The negative end of a water molecule is then able to easily and strongly attract to the hydrogen atom and pull it away

  3. Down a group, bond strength decreases • WHY?A weaker bond means that the hydrogen atom is more easily pulled away from the atom to which it is attached • Ie. HF is a stronger acid than water, but HF is the weakest of the hydrohalic acids (HCl, HI, HBr, etc.) because the H-F bond is relatively strong!

  4. Oxyacids • Acids that contain oxygen atoms • Increase in acid strength with increasing number of oxygen atoms • Oxygen is more electronegative than hydrogen, so oxygen atoms draw electrons away from hydrogen atoms • The more oxygen atoms there are in a molecule, the greater the polarity making it easier for the water molecule to tear the hydrogen atom away!

  5. When the central atoms are in the same group and have the same # of oxygen atoms, bondstrength increases/acid strength decreases from bottom to top within a group: • eg.) H2SO4(aq) H2SeO4(aq) >

  6. Oxyacids with the same # of oxygens on the central atom, acid strength increases from left to right in a period: • eg.) H2SO4(aq) H3PO4(aq) >

  7. For a given central atom, the acid strength of an oxyacid increases with the # of oxygens it holds: • eg.) H2SO4(aq) H2SO3(aq) • HNO3(aq) HNO2(aq) > >

  8. 8.2 Weak Acids & Bases

  9. Weak Acids • H2CO3(aq), HF(aq), CH3COOH(aq), H2CO3(aq), H2S(aq), H3PO3(aq), H3BO3(aq) • An acid that partially ionizes in solution but exists primarily in the form of molecules

  10. Percent Ionization • percent ionization (p) = • For the general weak acid ionization reaction, • HA(aq) H+(aq) + A-(aq) • [H+(aq)] = [HA(aq)]

  11. 1. In a 0.10 mol/L CH3COOH(aq) solution only 1.3% molecules ionize. Calculate the [H+(aq)]. • CH3COOH(aq)+ H+(aq) • [H+(aq)] = x 0.10 mol/L • = 1.3 x 10-3mol/L

  12. 2. The pH of a 0.050 mol/L methanoic acid solution is 2.78. Calculate the percent ionization of methanoic acid. • HCO2H(aq) H+(aq) + HCO2-(aq) • [HCO2H(aq)] = 0.050 mol/L • pH = 2.78 • [H+(aq)] = 10-pH • = 10-2.78= 1.7 x 10-3mol/L • p = x 100% • = x 100% • 3.3% • Methanoic acid ionizes 3.3% in a 0.050mol/L solution.

  13. Ionization Constants for Weak Acids • The equilbrium constant for weak acid is known as the acid ionization constant, Ka. • HA(aq) + H2O(l)A-(aq) + H3O+(aq) • Ka= • Ie. CH3COOH(aq)+ H+(aq) • Ka =

  14. 3. Calculate the Kaof hydrofluoric acid, HF(aq) , if a 0.100 mol/L solution at equilibrium at SATP has a percent ionization of 7.8% • HF(aq) H+(aq) + F-(aq) • Ka = • Let x be the change in concentration • x = 0.100mol/L x 0.078 • x = 0.0078 mol/L

  15. Ka = • = • Ka = 6.6 x 10-4 • We can compare the Kawith the values listed in Appendix C9… • The Ka of hydrofluoric acid at SATP is 6.6 x 10-4

  16. 4. Butyric acid, HC4H7O2, is used to make compounds employed in artificial flavourings and syrups. A 0.250 mol/L aqueous solution of HC4H7O2 is found to have a pH of 2.72. • HC4H7O2(aq) H+(aq) + C4H7O2-(aq) • .BUT.. x is not unknown, since we have the pH, we can find the [H+(aq)] • [H+(aq)] = 10-2.72= 1.9 x 10-3= x • Ka= = 1.47 x 10-5 • The Kaof the butyric acid solution is 1.47 x 10-5.

  17. Weak Bases • NH3, N2H4 (hydrazine) • A weak base has a weak attraction for protons • Base ionization constant, Kb • B(aq) + H2O(l) HB+(aq) + OH-(aq) • Kb = • ie. NH3(aq) + H2O(l) OH-(aq) + NH4+(aq) • Kb = Ionization Constants for Weak Bases

  18. The Relationship between Kaand Kb • For acids and bases whose chemical formulas differ only by a hydrogen, ie. Conjugate acid-base pairs, • KaKb = Kw • 5. What is the value of a base ionization constant, Kb for the acetate ion, C2H3O2-(aq) at SATP? (*HINT: appendix C9 has Ka values) • At SATP, Ka of acetic acid is 1.8 x 10-5 • KaKb = Kw • Kb = Kw / Ka = (1.0 x 10-14) / (1.8 x 10-5) • Kb = 5.6 x 10-10

  19. 6. Calculate the hydrogen ion concentration and the pH of a 0.10mol/L acetic acid solution. Ka for acetic acid is 1.8 x 10-5. • CH3COOH(aq)+ H+(aq) • Ka = • Ka= = 1.8 x 10-5 • The unknown can be solved (quadratic equation). Luckily, the calculation may be simplified 

  20. The calculation may be simplified by assuming that since Kais so small, CH3COOH(aq) will ionize very little and the value of x is expected to be very small. • Will need to use the hundred rule to dtermine whether this assumption is warranted.. • The Hundred Rule: if , a simplifying assumption can be made. • = 5.6 x 103 • Since 5.6 x 103 > 100,

  21. The equilibrium equation becomes.. • 1.8 x 10-5 • X21.8 x 10-6 • x = 1.3 x 10-3 • We must validate our assumption that 0.10 – x = 0.10. Since Kavalues are typically known to an accuracy 5%, in general, the approximation will be considered valid if, for the acid HA, • x = 1.3 x 10-3 , [HA]initial = 0.10 mol/L • = (1.3 x 10-3mol/L )/ (0.10 mol/L) x 100% = 1.3 % • Since 1.3% < 5%, the assumption we made was valid. Therefore..

  22. (last slide for this answer!!) • Therefore.. [H+(aq)] = x • [H+(aq)] = 1.3 x 10-3mol/L • pH = -log[H+(aq)] • = -log(1.3 x 10-3mol/L) • = 2.89

  23. 7. What is the pH of a solution that is 0.00250M CH3NH2(aq) ? For methylamine, Kb = 4.2 x 10-4.

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