Calculate the concentration of the species in excess and convert this value to pH. In the second step, we use the equilibrium equation to determine $$[\ce{H^{+}}]$$ of the resulting solution. Figure $$\PageIndex{1a}$$ shows a plot of the pH as 0.20 M $$\ce{HCl}$$ is gradually added to 50.00 mL of pure water. Conversely, for the titration of a weak base, where the pH at the equivalence point is less than 7.0, an indicator such as methyl red or bromocresol blue, with pKin < 7.0, should be used. $\ce{CH3CO2H(aq) + OH^{−} (aq) <=> CH3CO2^{-}(aq) + H2O(l)}$. Find the pH after each addition of acid from the buret. If we had added exactly enough hydroxide to completely titrate the first proton plus half of the second, we would be at the midpoint of the second step in the titration, and the pH would be 3.81, equal to $$pK_{a2}$$. Plots of acid–base titrations generate titration curves that can be used to calculate the pH, the pOH, the $$pK_a$$, and the $$pK_b$$ of the system. Before any base is added, the pH of the acetic acid solution is greater than the pH of the $$\ce{HCl}$$ solution, and the pH changes more rapidly during the first part of the titration. Now consider what happens when we add 5.00 mL of 0.200 M $$\ce{NaOH}$$ to 50.00 mL of 0.100 M $$CH_3CO_2H$$ (part (a) in Figure $$\PageIndex{3}$$). The ionization constant for the deprotonation of indicator $$\ce{HIn}$$ is as follows: $K_{In} =\dfrac{ [\ce{H^{+}} ][ \ce{In^{-}}]}{[\ce{HIn}]} \label{Eq3}$. If one species is in excess, calculate the amount that remains after the neutralization reaction. All problems of this type must be solved in two steps: a stoichiometric calculation followed by an equilibrium calculation. The pH tends to change more slowly before the equivalence point is reached in titrations of weak acids and weak bases than in titrations of strong acids and strong bases. Consider the titration of 300.0 mL of 0.500 M NH3 (Kb = 1.8 ×10-5) with 0.500 M HNO3. Then calculate the initial numbers of millimoles of $$OH^-$$ and $$CH_3CO_2H$$. Taking the negative logarithm of both sides, From the definitions of $$pK_a$$ and pH, we see that this is identical to. NH3 + HNO3 —–> NH4NO3 NH3 is a weak base and HNO3 is pretty strong acid. B The equilibrium between the weak acid ($$\ce{Hox^{-}}$$) and its conjugate base ($$\ce{ox^{2-}}$$) in the final solution is determined by the magnitude of the second ionization constant, $$K_{a2} = 10^{−3.81} = 1.6 \times 10^{−4}$$. Just as with the $$\ce{HCl}$$ titration, the phenolphthalein indicator will turn pink when about 50 mL of $$\ce{NaOH}$$ has been added to the acetic acid solution. Again we proceed by determining the millimoles of acid and base initially present: $100.00 \cancel{mL} \left ( \dfrac{0.510 \;mmol \;H_{2}ox}{\cancel{mL}} \right )= 5.10 \;mmol \;H_{2}ox$, $55.00 \cancel{mL} \left ( \dfrac{0.120 \;mmol \;NaOH}{\cancel{mL}} \right )= 6.60 \;mmol \;NaOH$. A dog is given 500 mg (5.80 mmol) of piperazine ($$pK_{b1}$$ = 4.27, $$pK_{b2}$$ = 8.67). In titrations of weak acids or weak bases, however, the pH at the equivalence point is greater or less than 7.0, respectively. One point in the titration of a weak acid or a weak base is particularly important: the midpoint of a titration is defined as the point at which exactly enough acid (or base) has been added to neutralize one-half of the acid (or the base) originally present and occurs halfway to the equivalence point. The value can be ignored in this calculation because the amount of $$CH_3CO_2^−$$ in equilibrium is insignificant compared to the amount of $$OH^-$$ added. They are typically weak acids or bases whose changes in color correspond to deprotonation or protonation of the indicator itself. Watch the recordings here on Youtube! This section describes a procedure called titration, which can be used to find the molarity of a solution of an acid or a base. Below the equivalence point, the two curves are very different. As shown in part (b) in Figure 17.4.3, the titration curve for NH3, a weak base, is the reverse of the titration curve for acetic acid. To calculate $$[\ce{H^{+}}]$$ at equilibrium following the addition of $$NaOH$$, we must first calculate [$$\ce{CH_3CO_2H}$$] and $$[\ce{CH3CO2^{−}}]$$ using the number of millimoles of each and the total volume of the solution at this point in the titration: $final \;volume=50.00 \;mL+5.00 \;mL=55.00 \;mL$ $\left [ CH_{3}CO_{2}H \right ] = \dfrac{4.00 \; mmol \; CH_{3}CO_{2}H }{55.00 \; mL} =7.27 \times 10^{-2} \;M$ $\left [ CH_{3}CO_{2}^{-} \right ] = \dfrac{1.00 \; mmol \; CH_{3}CO_{2}^{-} }{55.00 \; mL} =1.82 \times 10^{-2} \;M \nonumber$. answered  02/09/13, The pKb = -log(Kb) = -log(1.8 x 10^-5) = 4.7, As a side note, my book says the Kb is actually 1.74 x 10^-5. Dick B. Comparing the titration curves for $$\ce{HCl}$$ and acetic acid in Figure $$\PageIndex{3a}$$, we see that adding the same amount (5.00 mL) of 0.200 M $$\ce{NaOH}$$ to 50 mL of a 0.100 M solution of both acids causes a much smaller pH change for $$\ce{HCl}$$ (from 1.00 to 1.14) than for acetic acid (2.88 to 4.16). Thus the pH of a solution of a weak acid is greater than the pH of a solution of a strong acid of the same concentration. To calculate the pH at any point in an acid–base titration. To calculate the pH of the solution, we need to know $$\ce{[H^{+}]}$$, which is determined using exactly the same method as in the acetic acid titration in Example $$\PageIndex{2}$$: final volume of solution = 100.0 mL + 55.0 mL = 155.0 mL. As the equivalence point is approached, the pH drops rapidly before leveling off at a value of about 0.70, the pH of 0.20 M HCl. Adding only about 25–30 mL of $$\ce{NaOH}$$ will therefore cause the methyl red indicator to change color, resulting in a huge error. D We can obtain $$K_b$$ by substituting the known values into Equation \ref{16.18}: $K_{b}= \dfrac{K_w}{K_a} =\dfrac{1.01 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.80 \times 10^{-10} \label{16.23}$. 2) The point at which the pH = pKa, or pOH = pKb occurs at the half-equivalence point, (i.e., when enough titrant has been added to react with half of the acid or base present). The equivalence point in the titration of a strong acid or a strong base occurs at pH 7.0. The $$pK_{in}$$ (its $$pK_a$$) determines the pH at which the indicator changes color. As the acid or the base being titrated becomes weaker (its $$pK_a$$ or $$pK_b$$ becomes larger), the pH change around the equivalence point decreases significantly. Because the neutralization reaction proceeds to completion, all of the $$OH^-$$ ions added will react with the acetic acid to generate acetate ion and water: $CH_3CO_2H_{(aq)} + OH^-_{(aq)} \rightarrow CH_3CO^-_{2\;(aq)} + H_2O_{(l)} \label{Eq2}$. As explained discussed, if we know $$K_a$$ or $$K_b$$ and the initial concentration of a weak acid or a weak base, we can calculate the pH of a solution of a weak acid or a weak base by setting up a ICE table (i.e, initial concentrations, changes in concentrations, and final concentrations). Synthetic indicators have been developed that meet these criteria and cover virtually the entire pH range. The shape of a titration curve, a plot of pH versus the amount of acid or base added, provides important information about what is occurring in solution during a titration. Start here or give us a call: (312) 646-6365, © 2005 - 2020 Wyzant, Inc. - All Rights Reserved, a Question In contrast, the titration of acetic acid will give very different results depending on whether methyl red or phenolphthalein is used as the indicator. Both solutions were initially at 35.0 degrees celcius; the temperature of the resulting solution was recorded at 37.0 degrees celcius.