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pH, pKa, Ka, pKb, Kb

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Ranking Acid Base Strength Using Ka pKa Values Leah4sci

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Key Questions

  • The pH scale provides a way of measuring how acidic or basic solutions are. The scale ranges from 0-14. A pH of 0 is the most acidic, 7 is neutral and 14 is the most basic.

    Here is a video of a lab which looks at a number of different solutions and measures their pH levels using a pH meter and an indicator.

    video from: Noel Pauller

  • #"p"K_"a"# and #"p"K_"b"# are measures of the strengths of acids and bases, respectively

    Acids

    When you dissolve an acid in water, it undergoes an equilibrium reaction with the water in an.

    HA + H₂O ⇌ H₃O⁺ + A⁻

    The value of the equilibrium constant is given by

    #K_"a" = (["H"_3"O"^+]["A"^-]]/["HA"]#

    The greater the value of #K_"a"#, the stronger the acid.

    For most weak acids, #K_"a"# ranges from #10^-2# to #10^-14#.

    We convert these exponential numbers into a normal range by taking their negative logarithm.

    The operator #"p"# means "take the negative logarithm of".

    So #"p"K_"a" = -logK_"a"#.

    For most weak acids, #"p"K_"a"# ranges from 2 to 14.

    Thus, the smaller the value of #"p"K_"a"# , the stronger the acid.

    Bases

    When you dissolve a base in water, it reacts with the water in an equilibrium reaction.

    B + H₂O ⇌ BH⁺ + OH⁻

    The value of the equilibrium constant is given by

    #K_"b" = (["BH"^+]["OH"^-]]/["B"]#

    The greater the value of #K_"b"#, the stronger the base.

    For most weak acids, #K_"b"# ranges from #10^-2# to #10^-13#.

    #"p"K_"b" = -logK_"b"#.

    For most weak acids, #"p"K_"a"# ranges from 2 to 13.

    The smaller the value of #"p"K_"b"# , the stronger the base.

    Here's a video on #"p"K_"a"# and #"p"K_"b"#.

  • Answer:

    These are measures of acidity and basicity...

    Explanation:

    And acid in aqueous solution is conceived to undergo a protonolysis reaction...

    #HX(aq) + H_2O(l) rightleftharpoonsH_3O^+ + X^(-)#

    And as for any equilibrium, we can measure and quantify it in the usual way...

    #K_a=([H_3O^+][X^-])/([HX(aq)])#

    Note that #H_2O# DOES NOT appear in the equilibrium expression because it is present in such high concentration that it is effectively constant..

    For strong acids, i.e. #HI#, #HBr#, #HCl#, #H_2SO_4# protonolysis is effectively quantitative: the given equilibrium lies entirely to the right as we face the page, and the acid solution is quantitative in #H_3O^+#. For weaker acids, #HF#, #H_3C-CO_2H#, the equilibrium lies somewhat to the left...and concentrations of the parent acid remain at equilibrium.

    And likewise, we can formalize the performance of a base by an equivalent equilibrium...we use ammonia, because this is a WEAK base in aqueous solution...

    #NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + HO^-#

    And #K_b# is defined in an equivalent way to #K_a#...

    #K_b=([NH_4^+][HO^-])/([NH_3(aq)])#, #K_b"(ammonia)"=1.74xx10^-5#...

    Confused yet....?

    Well, note that NECESSARILY....for a given acid/conjugate pair, say #NH_4^+"/"NH_3#...

    #K_aK_b=10^-14#...or perhaps more usefully...

    #pK_a+pK_b=14#...

Questions

  • il maestro answered · 4 months ago