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Now consider the titration of a weak acid, HA, with NaOH titrant. The neutralization reaction isceHA + OH^– rightarrow A^– + H_2OThere are four distinct regions in the course of such a titration with four different approaches to calculating the pH in each region. To illustrate this general approach, consider the titration of 10.0 mL of (0.0300 ,M ,textbenzoic acid) (the analyte) with (0.0100, M, ceNaOH(aq)) solution (the titrant). Benzoic acid has the formula (ceC6H5CO2H), but for simplicity we will abbreviate it as (ceHBz), and it has a (K_a = 6.46 times 10^–5). Calculate the pH at each of the following points in the titration: the initial point, before any titrant has been added; after adding 10.0 mL of 0.0100 M NaOH(aq); after adding 15.0 mL of 0.0100 M NaOH(aq); after adding 30.0 mL of 0.0100 M NaOH(aq); after adding 45.0 mL of 0.0100 M NaOH(aq). Once again, before carrying out any other calculations, determine (a) the number of millimoles of analyte initially present, and (b) the volume of titrant that must be added to reach the equivalence point. It will be useful to keep these numbers in mind as we proceed to calculate the pH values throughout the course of the titration. The number of millimoles of HBz initially present is textmmol, ceHBz = (0.0300, M)(10.0, M) = 0.300 ,textmmol For our example titration the neutralization reaction is ceHBz(aq) + NaOH(aq) <=> H2O(l) + NaBz(aq) At the equivalence point textmmol, ceHBz,textinitially present = textmmol, ceNaOH, textadded Therefore, the volume of (ceNaOH(aq)) solution needed to reach the equivalence point is V_NaOH = dfracM_HBzV_HBzM_NaOH = dfrac(0.0300,M)(10.0,mL)0.0100,M = 30.0,mL Note that these are the same values we obtained in the example (HBrNaOH) titration above, and we will be calculating the pH at the some of the same points throughout the course of the titration. You will want to compare the procedures and results in this weak acid - strong base example to those of the previous strong acid - strong base example. (i) The initial point, before any titrant has been added Here we simply have a solution of (0.0300, M ,ceHBz). We must use (K_a = 6.46 times 10^–5) to calculate (ceH3O^+) and pH. But in this case, (C_HBz gg K_a), so we can ignore hydrolysis of molecular (HBz) and use the approximate method to calculate (ceH3O^+), without needing to solve the quadratic equation. ceH3O^+ = sqrtC_HBz K_a = sqrt(0.0300)(6.46 times 10^-5) = 1.39 times 10^-3,M therefore pH = 2.856 (ii) After adding 10.0 mL of 0.0100 M NaOH(aq) The number of millimoles of added NaOH is textmmol, ceNaOH = (0.0100 ,M)(10.0, mL) = 0.100,textmmol We can summarize the numbers of millimoles of analyte and titrant before and after the addition by the following pseduo-ICE table: ICE Table (ceHBz(aq)) (ceNaOH) (ce H2O(l)) (ceNaBz(aq)) Initial (0.300, mmol) (0.100, mmol) - 0 After Reaction (0.200, mmol) 0 - (0.100, mmol) This is a buffer solution, so we really do not need to be concerned with the total volume of the solution at this point, because a buffers pH is independent of volume. All we need to do is substitute the numbers of millimoles of HBz and Bz– into the Ka expression and solve for (ceH3O^+). K_a = cedfracH3O^+Bz^-HBz = dfracceH3O^+(0.100, mmol)0.300, mmol = 6.46 times 10^-5 ceH3O^+ = 1.292 times 10^–4, M therefore pH = 3.889 (iii) After adding 15.0 mL of 0.0100 M NaOH(aq) The number of millimoles of added NaOH is textmmol, ceNaOH = (0.0100, M)(15.0, mL) = 0.150, textmmol We can summarize the numbers of millimoles of analyte and titrant before and after the addition by the following table: ICE Table (ceHBz(aq)) (ceNaOH) (ce H2O(l)) (ceNaBz(aq)) Initial (0.300, mmol) (0.150, mmol) - 0 After Reaction (0.150, mmol) 0 - (0.150, mmol) Note that we have added half the volume of (ceNaOH(aq)) solution needed to reach the equivalence point. This is called the half-titration point. As the table above shows, at the half-titration point, the solution contains equal numbers of millimoles of acid and conjugate base, making it an equimolar buffer solution. If we substitute into the (K_a) expression we obtain K_a = cedfracH3O^+Bz^-HBz = dfracceH3O^+(0.150, mmol)0.150, mmol = 6.46 times 10^-5 From which it follows pH = pK_a = –log_10(6.46 times 10^–5) = 4.190 This is a general result worth knowing: At the half-titration point in any weak acid strong base titration, pH = pKa (iv) After adding 30.0 mL of 0.0100 M NaOH(aq) The number of millimoles of added NaOH is textmmol, ceNaOH, textadded = (0.0100 ,M)(30.0, mL) = 0.300, textmmol This is the equivalence point (see preliminary calculation above), because we have added as many millimoles of NaOH titrant as there were millimoles of HBz analyte in the sample. At this point all of the HBz has been converted to Bz– , the conjugate base ICE Table (ceHBz(aq)) (ceNaOH) (ce H2O(l)) (ceNaBz(aq)) Initial (0.300, mmol) (0.300, mmol) - 0 After Reaction 0 0 - (0.300, mmol) We now have a solution of 0.300 mmol pure weak base (ceBz^–) in a volume that is the sum of the initial volume plus the volume of added titrant; i.e., V_tot = V_HBz + V_NaOH = 10.0 ,mL + 30.0, mL = 40.0, mL and C_Bz = dfrac0.300, mmol40.0, mL = 7.50 times 10^–3, M The pH of the solution will be determined by the hydroxide ion produced by the hydrolysis of the weak base (ceBz^–). ceBz^– + H2O <=> HBz + OH^– (Note that this is the back reaction to the neutralization reaction.) To calculate (ceOH^–), we need the (K_b) of (ceBz^–), which can be obtained from the relationship K_b^Bz^- = dfracK_wK_a^HBz = dfrac1.00 times 10^-146.46 times 10^-5 = 1.548 times 10^-10 As this value indicates, Bz– is a very weak base, so we can ignore loss of the analytical concentration from hydrolysis and solve for (ceOH^–) without using the quadratic equation. ce OH^- = sqrt C_Bz^-K_b =- sqrt(7.50 times 10^-3)(1.548 times 10^-10) = 1.077 times 10^-6,M therefore beginalign pOH &= 5.968 pH &= 8.032 endalign Note that the pH is greater than 7, as expected for a weak acid - strong base titration. (iv) After adding 45.0 mL of 0.0100 M NaOH(aq) This is 15.0 mL beyond the equivalence point. The number of millimoles of added NaOH is textmmol, ceNaOH, textadded = (0.0100 ,M)(45.0, mL) = 0.450 ,mmol At this point our table of millimoles of analyte and titrant before and after the addition now has the following values: ICE Table (ceHBz(aq)) (ceNaOH) (ce H2O(l)) (ceNaBz(aq)) Initial (0.300, mmol) (0.450, mmol) - 0 After Reaction 0 (0.150, mmol) - (0.450, mmol) The solution contains 0.150 mmol of (ceNaOH) beyond what was needed to neutralize the 0.300 mmol of (ceHBz) initially present. This amount of excess (ceOH^–) is vastly greater than any amount supplied by the weak base (ceBz^–), and it alone governs the pOH and pH of the solution. In other words, the calculation in this region is identical to the previous strong acid - strong base case. As before, the total volume of the solution at this point is V_tot = V_HBr + V_NaOH = 10.0, mL + 45.0, mL = 55.0, mL Therefore, OH– = CNaOH = 0.150 mmol55.0 mL = 2.727 times 10^–3, M pOH = 2.564 pH = 11.436 Consider 25.0 mL of (0.100, M, ceHA), for which (K_a = 1.00 times 10^–5), titrated with (0.0500 ,M , ceNaOH) solution. How many milliliters of (ceNaOH(aq)) are needed to reach the equivalence point? How many millimoles of (ceHA) are initially present in the sample? What is the initial pH, before adding any titrant? What is the pH after adding 10.0 mL of titrant? What is the pH after adding 25.0 mL of titrant? What is the pH after adding 50.0 mL of titrant? What is the pH after adding 75.0 mL of titrant?

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