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Acid Base Equilibrium

Acid Base Equilibrium. Weak Acids & Bases. Recall From Yesterday…. pH = -log [H 3 O + ] [H 3 O + ] = 10 -pH pOH = -log [OH - ] [OH - ] = 10 -pH pK w = pH + pOH = 14. Weak Acids. Donate only one H + Do not dissociate at 100% For a Weak Acid:

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Acid Base Equilibrium

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  1. Acid Base Equilibrium Weak Acids & Bases

  2. Recall From Yesterday…. pH = -log [H3O+] [H3O+] = 10-pH pOH = -log [OH-] [OH-] = 10-pH pKw= pH + pOH = 14

  3. Weak Acids • Donate only one H+ • Do not dissociate at 100% For a Weak Acid: The higher the Ka, the stronger the acid

  4. Acid Dissociation Constant • Ka • – also known as the acid ionization constant.

  5. pH and Ka of a Weak Acid • The smaller the value of Ka, the less the acid ionizes in aqueous solution.

  6. Ka constants

  7. 1. % Ionization • the fraction of acid molecules that dissociate compared with the initial concentration of the acid, expressed as a percent. • depends on the value of Ka for the acid, as well as the initial concentration of the weak acid. • For a Weak Acid: Or

  8. Try This: A 0.25M solution of HF(aq) is ionized at 21.3%, calculate the pH

  9. % Dissociation % dissociation = [HA]dissociated x 100% [HA] initial

  10. 2. pH of a Weak Acid, Given the Ka • 1. Balanced equation for acid equilibrium • 2. Equilibrium constant (Ka) expression • 3. ICE Table

  11. pH of a Weak Acid, Given the Ka • 4. Substitute the [equilibrium] into the Ka equation. • 5. Solve for x i) 100 rule ii) quadratic equation • 6. Calculate pH from [H+]

  12. Try This: Chloracetic acid, is a weak acid (Ka = 1.36 x 10-3). Determine the pH of a 12.0 M solution of chloracetic acid.

  13. 3. Ka, Given the pH and [HA] Step 1: [H+] = 10-pH Step 2: [A-] = [H+] Step 3:

  14. Try This: You measure the pH of a 0.10 M hypochlorous acid solution, HOCl(aq) and find it to be 4.23. What is the Ka for hypochlorous acid?

  15. 4. Polyprotic acids • A polyprotic acid is capable of donating more than one proton (H2CO3(aq), H2SO4(aq)) • There is an ionization constant for each proton donation (Ka1, Ka2, etc.) as the ionization occurs in steps. (Table Pg. 803) • The Ka values become smaller with each ionization step, as the removal of a proton from a negatively charged object becomes more difficult.

  16. 4. Polyprotic acids

  17. Polyprotic acids Polyprotic acids • However, the most ionization occurs in the first step. • Ka1>> Ka2 > Ka3 .. . . • Consequently, the [H+] is predominantly established in the first ionization with the Ka1 value. Subsequent ionizations (Ka2 & Ka3) only add minimal amounts of [H+]. • Use Ka1 to determine the pH of the solution at equilibrium.

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