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7a. pKa and pKb of conjugate acids and bases

http://www.chemistry.jamesmungall.co.uk Acid Base Chemistry 7. Relative acidity and basicity -- competition for H+ a. pKa and pKb of conjugate acids and bases [slide 1] Now we're going to relate Ka and Kb values for conjugate acids and bases. So, we are learning to work out pKb from pKa and work out pKa from pKb, using the relation pKa + pKb equals 14. Then we can use this to calculate the pH for conjugate acids and bases. [slide 2] Let's remind ourselves what we mean by conjugate acids and bases. If we have a weak acid, HA, then if that breaks apart into A- and H+, the A- is called the conjugate base, and it will be a weak base, because this reaction can go backwards. [slide 3] The kind of question we want to answer is, if we're given the Ka or pKa value of a weak acid, in this case phenol, what will be the pH of a solution containing not the weak acid, but the conjugate base. [slide 4] For example, what is the pH of 0.1moldm-3 sodium phenoxide? Since the phenoxide ion is a weak base, we want to know the Kb value for this ion reacting with water to form phenol and hydroxide ions. Well, we've been given pKa for the conjugate acid, and we can use the relation pKa + pKb =14 and rearrange this to make pKb the subject, which gives pKb equals 14 minus pKa, which is this case is 14 - 9.9 which equals 4.1 [slide5] Now, this has been simplified to a weak base question, because we know the concentration of the weak base, 0.1moldm-3, and we know Kb, or at pKb which can easily be converted to Kb. To calculate the pH of a weak base, we say hydroxide concentration equals the square root of Kb multiplied by the base concentration, in this case that it the square root of 10-4.1 - that's converted our pKb value into a Kb value, just like we would do to convert pH and H+ concentration - multiplied by 0.1 , which gives 2.88 x 10 -3 moldm-3. pOH is then minus log of this which gives 2.6, and pH is 14 minus this which is 11.4. This should seem like a reasonable answer, since we have calculated the pH of a weak base, which you would expect to be in the region between 7 and 14. [slide 6] We can do these calculations the other way around, starting from a weak base. If a weak base, B, is protonated to form BH+, then BH+ will be a weak acid. [slide 7] For example, NH3 in equilibrium with NH4+ and OH- has a pKb value of 4.8. [slide 8] Now, what is the pH of a 0.1moldm-3 solution of NH4+Cl-, ammonium chloride? We really want to know the Ka value of the revervse reaction, NH4+ behaving as a weak acid, and partially dissociating into NH3 and H+. Once again, we can use the relation pKa + pKb equals 14, and rearrange this as pKa equals 14 minus pKb, which in this case is 14 minus 4.8 which equals 9.2. [slide 9] Continuing with this, we now have a weak acid question in which we know the concentration of the weak acid and the pKa. Therefore, we can work out the H+ concentration by taking the square root of the Ka multiplied by the AH concentration, which in this case is square root of 10 to minus 9.2, converting the pKa into a Ka value, multiplied by 0.1, which gives 7.94 x 10-6 moldm-3. pH is then minus log of this, which is 5.1. Once again, this is a sensible answer, since it lies between 0 and 7, meaning that it is a somewhat acidic solution. [slide 10] It's not really fair of me just to give you the formula pKa + pKb equals 14, so here's the proof. For HA dissociating, we can write an expression for Ka which is H+ times A- divided by HA concentration. For the conjugate base, A-, we can write an expression for Kb, which is AH times OH minus divided by A-. Then if we multiply these two equilibrium constants together we have H+ times A- divided by HA, multiplied by AH times OH- divided by A-. The AH and A- terms both cancel out, so this expression simplifies to H+ concentration multiplied by the OH- concentration which is exactly how we define Kw, the ionic product of water.
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