Titration Of H3po4 With Naoh
Out of Course Consignment #4, Problem 2
Starting with thirty ml of 0.1 Chiliad citric acid, calculate the initial pH and the pH at each v ml increase of 0.1 Thousand NaOH until you are x ml past the concluding equivalence signal. Plot the data and determine whether 99.nine% of the citric acid has been neutralized at the last equivalence indicate. Also calculate the concentration of all species in solution at the second equivalence point.
It is worth examining this problem in some particular since we accept not done anything exactly like it in class. Essentially it consists of the titration of a polyprotic acid using a stiff base of operations. Citric acrid is a mutual buffer but is an interesting example because the get-go two pKa values are fairly shut to each other.
If we look in the tabular array nosotros notice out that citric acid (H3cit) is a triprotic acid. The post-obit iii equilibrium reactions define the system.
\[\begin{align}
&\ce{H3cit + H2o \leftrightarrow H2cit- + H3O+} &&\mathrm{K_{a1} = 7.45\times 10^{-4}}\\
&\ce{H2cit^2- + H2O \leftrightarrow Hcit^ii- + H3O+} &&\mathrm{K_{a2} = 1.73\times 10^{-5}}\\
&\ce{Hcit- + Water \leftrightarrow cit^3- + H3O+} &&\mathrm{K_{a3} = 4.02\times ten^{-7}}
\end{marshal}\]
Fifty-fifty though the start two Ka values are adequately close to each other, nosotros tin can still use only the Ka1 expression to solve for the initial pH.
\[\begin{align}
& &&\ce{H3cit}\hspace{25px} + \hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{H2cit-} \hspace{25px} + &&\ce{H3O+} &&\mathrm{K_{a1}}\\
&\ce{Initial} &&0.1 &&0 &&0 && \\
&\ce{Equilibrium} &&0.1 - \ce{10} &&\ce{x} &&\ce{x} && \\
&\ce{Approximation} &&0.1 &&\ce{x} &&\ce{ten} &&
\stop{align}\]
\[\mathrm{K_{a1}=\dfrac{[H_2cit^-][H_3O^+]}{[H_3cit]}=\dfrac{(x)(x)}{0.i}=vii.45\times10^{-4}}\]
\[\mathrm{10 = [H_3O^+] = 0.00863 \hspace{60px} pH = ii.06}\]
If we check the approximation, it actually turns out that the value is too loftier and that we should take used a quadratic if we wanted the exact respond. But the value of 2.06 volition suffice for now.
The side by side step is to consider what happens when nosotros commencement adding sodium hydroxide to the solution. This will catechumen Hiiicit into the other forms, and we can start the procedure past assuming it will occur in a stepwise mode (in other words, H3cit will be converted into Htwocit– by the base of operations until all the Hthreecit is used upwardly, then H2cit– will be converted into Hcitii–, etc.). This would allow us to construct the chart shown in Table 3 of the moles of each species that would occur over the course of the titration.
Table three. Moles of each species in the titration of citric acid (0.1 M, 30 ml) with NaOH (0.1M).
| ml NaOH | Hthreecit | H2cit– | Hcit2– | citthree– | pH |
|---|---|---|---|---|---|
| 0 | 0.0030 mol | 0 | 0 | 0 | 2.06 |
| five | 0.0025 | 0.0005 | 0 | 0 | |
| x | 0.0020 | 0.0010 | 0 | 0 | |
| 15 | 0.0015 | 0.0015 | 0 | 0 | |
| 20 | 0.0010 | 0.0020 | 0 | 0 | |
| 25 | 0.0005 | 0.0025 | 0 | 0 | |
| thirty | 0 | 0.0030 | 0 | 0 | |
| 35 | 0 | 0.0025 | 0.0005 | 0 | |
| forty | 0 | 0.0020 | 0.0010 | 0 | |
| 45 | 0 | 0.0015 | 0.0015 | 0 | |
| l | 0 | 0.0010 | 0.0020 | 0 | |
| 55 | 0 | 0.0005 | 0.0025 | 0 | |
| 60 | 0 | 0 | 0.0030 | 0 | |
| 65 | 0 | 0 | 0.0025 | 0.0005 | |
| 70 | 0 | 0 | 0.0020 | 0.0010 | |
| 75 | 0 | 0 | 0.0015 | 0.0015 | |
| 80 | 0 | 0 | 0.0010 | 0.0020 | |
| 85 | 0 | 0 | 0.0005 | 0.0025 | |
| ninety | 0 | 0 | 0 | 0.0030 | |
| 95 | 0 | 0 | 0 | 0.0030 | |
| 100 | 0 | 0 | 0 | 0.0030 |
If we examine the increments from 5 ml to 25 ml, nosotros run across that we accept observable quantities of H3cit and Htwocit–, which are both members of a cohabit pair. This region is a buffer solution and the pH can exist adamant using Ka1.
\(\textrm{5 ml:} \hspace{32px} \mathrm{pH=pK_{a1}+ \log\left( \dfrac{[H_2cit^-]}{[H_3cit]}\right )=iii.128+\log\left( \dfrac{0.0005\: mol}{0.0025\: mol}\right )=two.43}\)
\(\textrm{x ml:}\hspace{25px} \mathrm{pH=pK_{a1}+ \log\left( \dfrac{[H_2cit^-]}{[H_3cit]}\correct )=3.128+\log\left( \dfrac{0.0010\: mol}{0.0020\: mol}\right )=2.83}\)
\(\textrm{15 ml:} \hspace{25px} \mathrm{pH = pK_{a1}+ \log\left( \dfrac{[H_2cit^-]}{[H_3cit]}\right ) = 3.128 + \log\dfrac{0.0015\: mol}{0.0015\: mol} = iii.128}\)
Note that the pH at this increase is pKa1.
\(\textrm{20 ml:} \hspace{25px} \mathrm{pH=pK_{a1}+ \log\left( \dfrac{[H_2cit^-]}{[H_3cit]}\right )=3.128+\log\left( \dfrac{0.0020\: mol}{0.0010\: mol}\right )= iii.43}\)
\(\textrm{25 ml:} \hspace{25px} \mathrm{pH = pK_{a1}+ \log\left( \dfrac{[H_2cit^-]}{[H_3cit]}\right ) = iii.128 + \log\left( \dfrac{0.0025\: mol}{0.0005\: mol}\right ) = 3.83}\)
30 ml: This is the first equivalence point, since we have converted all of the H3cit to H2cit–. At this indicate we have "all" of the commencement intermediate (Hiicit–) and tin can calculate the pH using the expression (pK1 + pK2)/2
\[\mathrm{pH = \dfrac{pK_{a1}+ pK_{a2}}{2}=\dfrac{iii.128+iv.761}{two}=3.9}\]
If we examine the region from 35 to 55 ml, we have appreciable quantities of H2cit– and Hcit2–, a buffer solution based on Ka2.
\(\textrm{35 ml:} \hspace{25px} \mathrm{pH = pK_{a2}+ \log\dfrac{[Hcit^{2-}]}{[H_2cit^-]} = four.761 + \log\left( \dfrac{0.0005\: mol}{0.0025\: mol}\correct ) = 4.06}\)
\(\textrm{40 ml:}\hspace{25px} \mathrm{pH = pK_{a2}+ \log \left(\dfrac{[Hcit^{2-}]}{[H_2cit^-]}\correct) = four.761 + \log \left( \dfrac{0.0010\: mol}{0.0020\: mol}\right ) = four.46}\)
\(\textrm{45 ml:}\hspace{25px} \mathrm{pH = pK_{a2}+ \log \left(\dfrac{[Hcit^{two-}]}{[H_2cit^-]}\right) = 4.761 + \log \left( \dfrac{0.0015\: mol}{0.0015\: mol}\right ) = 4.761}\)
Notation that at this point, the pH is equal to pKa2.
\(\textrm{50 ml:}\hspace{25px} \mathrm{pH = pK_{a2}+ \log \left(\dfrac{[Hcit^{ii-}]}{[H_2cit^-]}\right) = four.761 + \log \left( \dfrac{0.0020\: mol}{0.0010\: mol}\right ) = 5.06}\)
\(\textrm{55 ml:}\hspace{25px} \mathrm{pH = pK_{a2}+ \log \left(\dfrac{[Hcit^{2-}]}{[H_2cit^-]}\right) = iv.761 + \log \left( \dfrac{0.0025\: mol}{0.0005\: mol}\right ) = 5.46}\)
60 ml: This is the second equivalence point, since nosotros have converted all of the Hiicit- to Hcit2-. At this point we have "all" of the 2nd intermediate (Hcitii-) and can calculate the pH using the expression (pK2 + pK3)/2
\[\mathrm{pH = \dfrac{pK_{a2}+ pK_{a3}}{2}=\dfrac{iv.761 + six.396}{ii}= 5.58}\]
If nosotros examine the region from 65 to 85 ml, nosotros accept appreciable quantities of Hcit2- and cit3-, a buffer solution based on Ka3.
\(\textrm{65 ml:}\hspace{25px} \mathrm{pH = pK_{a3}+ \log \left(\dfrac{[cit^{3-}]}{[Hcit^{2-}]}\right) = 6.396 + \log \left( \dfrac{0.0005\: mol}{0.0025\: mol}\right ) = 5.lxx}\)
\(\textrm{seventy ml:}\hspace{25px} \mathrm{pH = pK_{a3}+ \log \left(\dfrac{[cit^{three-}]}{[Hcit^{2-}]}\right) = six.396 + \log \left( \dfrac{0.0010\: mol}{0.0020\: mol}\correct ) = 6.09}\)
\(\textrm{75 ml:}\hspace{25px} \mathrm{pH = pK_{a3}+ \log \left(\dfrac{[cit^{3-}]}{[Hcit^{ii-}]}\right) = 6.396 + \log \left( \dfrac{0.0015\: mol}{0.0015\: mol}\right ) = 6.396}\)
Note that at this betoken, the pH is equal to pKa3.
\(\textrm{80 ml:}\hspace{25px} \mathrm{pH = pK_{a3}+ \log \left(\dfrac{[cit^{three-}]}{[Hcit^{2-}]}\right) = 6.396 + \log \left( \dfrac{0.0020\: mol}{0.0010\: mol}\correct ) = 6.lxx}\)
\(\textrm{85 ml:}\hspace{25px} \mathrm{pH = pK_{a3}+ \log \left(\dfrac{[cit^{three-}]}{[Hcit^{2-}]}\right) = 6.396 + \log \left( \dfrac{0.0025\: mol}{0.0005\: mol}\correct ) = 7.09}\)
xc ml: This is the 3rd equivalence point, since we have converted all of the Hcit2– to cit3–. To a get-go approximation we simply have cit3– in solution. This is a polybasic base, but equally we have done earlier, we only demand to consider the outset reaction in the series to calculate the pH. The relevant reaction, which is the Kb value of 1000a3, is shown below.
citiii– + HiiO = Hcit2– + OH– Gb of Ka3 = 2.5×10-8
Nosotros need to calculate the concentration of cit3– that is present in solution, recognizing that the titrant caused a dilution of the initial concentration of citric acid (thirty ml of initial solution and xc ml of additional titrant).
\[\textrm{Molarity of citrate = (0.0030 mol/0.120 L) = 0.025 M}\]
\[\begin{align}
& &&\ce{cit^three-}\hspace{25px} + \hspace{25px} \ce{Water} \hspace{25px}\leftrightarrow &&\ce{Hcit^2-} \hspace{25px} + &&\ce{OH-} \\
&\ce{Initial} &&0.025 &&0 &&0 \\
&\ce{Equilibrium} &&0.025 - \ce{ten} &&\ce{x} &&\ce{x} \\
&\ce{Approximation} &&0.025 &&\ce{x} &&\ce{10}
\cease{marshal}\]
\[\mathrm{K_{b3} =\dfrac{[Hcit^{2-}][OH^-]}{[cit^{three-}]}=\dfrac{(x)(x)}{0.025} = two.5\times10^{-8}}\]
\[\mathrm{x = [OH^-] = 2.5\times ten^{-5}}\]
\[\mathrm{pOH = four.half dozen \hspace{60px} pH = ix.iv}\]
Checking the approximation shows that it was valid in this instance. Note that the pH at this equivalence point is basic, which is non surprising since cit3– is a base.
\[\dfrac{ii.five×10^{-5}}{0.025}× 100 = 0.i\%\]
95 ml: In this example we take a mixture of a strong base (NaOH) with a weaker base (citrate). The extra amount of potent base of operations (v ml or 0.0005 moles) will determine the pH.
\[\mathrm{[OH^-] =\dfrac{0.0005\: mol}{0.125\: L} = four.0\times10^{-3}}\]
\[\mathrm{pOH = 2.4 \hspace{60px} pH = 11.6}\]
100 ml: Once again, the pH is determined by the amount of actress strong base present in the solution (x ml or 0.0010 moles).
\[\mathrm{[OH^-] =\dfrac{0.0010\: mol}{0.130\: L} = 7.7\times10^{-three}}\]
\[\mathrm{pOH = two.i \hspace{60px} pH = 11.9}\]
We tin at present compile an entire chart (Table 4) of the changes that occur over this titration:
Table iv. Calculated pH values for the titration of citric acrid (0.i M, 30 mL) with NaOH (0.1 G).
| ml NaOH | H3cit | H2cit– | Hcittwo– | cit3– | pH |
|---|---|---|---|---|---|
| 0 | 0.0030 m | 0 | 0 | 0 | two.06 |
| 5 | 0.0025 | 0.0005 | 0 | 0 | 2.43 |
| ten | 0.0020 | 0.0010 | 0 | 0 | 2.83 |
| 15 | 0.0015 | 0.0015 | 0 | 0 | 3.128 (pKa1) |
| xx | 0.0010 | 0.0020 | 0 | 0 | 3.43 |
| 25 | 0.0005 | 0.0025 | 0 | 0 | 3.83 |
| 30 | 0 | 0.0030 | 0 | 0 | 3.94 (pKa1+pKa2)/2 |
| 35 | 0 | 0.0025 | 0.0005 | 0 | 4.06 |
| 40 | 0 | 0.0020 | 0.0010 | 0 | 4.46 |
| 45 | 0 | 0.0015 | 0.0015 | 0 | iv.761 (pKa2) |
| l | 0 | 0.0010 | 0.0020 | 0 | five.06 |
| 55 | 0 | 0.0005 | 0.0025 | 0 | 5.46 |
| 60 | 0 | 0 | 0.0030 | 0 | 5.58 (pKa2+pKa3)/2 |
| 65 | 0 | 0 | 0.0025 | 0.0005 | five.70 |
| lxx | 0 | 0 | 0.0020 | 0.0010 | 6.09 |
| 75 | 0 | 0 | 0.0015 | 0.0015 | 6.396 (pKa3) |
| lxxx | 0 | 0 | 0.0010 | 0.0020 | half dozen.70 |
| 85 | 0 | 0 | 0.0005 | 0.0025 | 7.09 |
| 90 | 0 | 0 | 0 | 0.0030 | 9.40 |
| 95 | 0 | 0 | 0 | 0.0030 | xi.60 |
| 100 | 0 | 0 | 0 | 0.0030 | 11.90 |
It is especially helpful to plot these values versus the ml of titrant every bit shown in Effigy 3.
Figure 3. pH versus ml titrant for the titration of citric acrid (0.1 M, xxx mL) with NaOH (0.one One thousand).
At that place are several things worth noting in this plot. One is the way that the first ii equivalence points alloy together and there are no clear breaks in the plot. The only equivalence signal in this titration that is readily observable is the third. The other is to note that citric acid has a significant buffer region that stretches from a pH of nearly 2.v to 5.5. Citric acrid is normally used as a buffer for this pH region.
It is likewise worth examining what would be observed for a similar plot of a different triprotic acid. The information in Table 5 is for an identical titration of phosphoric acid.
Table 5. pH values for the titration of phosphoric acid (0.i M, xxx ml) with NaOH (0.one Chiliad).
| ml NaOH | H3PO4 | \(\ce{H2PO4-}\) | \(\ce{HPO4^2-}\) | \(\ce{PO4^3-}\) | pH |
|---|---|---|---|---|---|
| 0 | 0.0030 mol | 0 | 0 | 0 | - |
| v | 0.0025 | 0.0005 | 0 | 0 | 1.45 |
| 10 | 0.0020 | 0.0010 | 0 | 0 | i.85 |
| 15 | 0.0015 | 0.0015 | 0 | 0 | ii.148 (pKa1) |
| twenty | 0.0010 | 0.0020 | 0 | 0 | 2.45 |
| 25 | 0.0005 | 0.0025 | 0 | 0 | 2.85 |
| 30 | 0 | 0.0030 | 0 | 0 | 4.673 (pKa1+pKa2)/ii |
| 35 | 0 | 0.0025 | 0.0005 | 0 | six.l |
| 40 | 0 | 0.0020 | 0.0010 | 0 | vi.xc |
| 45 | 0 | 0.0015 | 0.0015 | 0 | 7.198 (pKa2) |
| 50 | 0 | 0.0010 | 0.0020 | 0 | 7.50 |
| 55 | 0 | 0.0005 | 0.0025 | 0 | 7.xc |
| sixty | 0 | 0 | 0.0030 | 0 | 9.789 (pKa2+pKa3)/two |
| 65 | 0 | 0 | 0.0025 | 0.0005 | 11.68 |
| lxx | 0 | 0 | 0.0020 | 0.0010 | 12.08 |
| 75 | 0 | 0 | 0.0015 | 0.0015 | 12.38 (pKa3) |
| fourscore | 0 | 0 | 0.0010 | 0.0020 | 12.68 |
| 85 | 0 | 0 | 0.0005 | 0.0025 | - |
| 90 | 0 | 0 | 0 | 0.0030 | - |
| 95 | 0 | 0 | 0 | 0.0030 | 11.60 |
| 100 | 0 | 0 | 0 | 0.0030 | xi.xc |
It is worth realizing that a few data points have been omitted since there is a problem at the beginning and over again at almost 75 ml of titrant. In the early on function, the acrid is strong enough that a adequately significant proportion dissociates. At the latter part of the titration, the base is and so strong that we really do not catechumen all of the \(\ce{HPO4^2-}\) to \(\ce{PO4^iii-}\) every bit implied. Even with this trouble, nosotros can examine a generalized plot for the titration of phosphoric acid with sodium hydroxide (Figure 4).
Figure iv. pH versus ml titrant for the titration of phosphoric acid (0.1 Thou, xxx mL) with NaOH (0.ane M).
Note hither that the get-go two equivalence points are obvious, whereas the third equivalence indicate will non exist distinguishable because of the very high pKa3 value. The concentration of citric or phosphoric acrid can exist determined through a titration with sodium hydroxide, provided you realize which equivalence points tin can be successfully monitored during the titration.
Now we can examine the last two parts of the homework problem. The outset is whether 99.ix% of the species is in the form cit3– at the third equivalence point. Going back to the calculation at 90 ml of titrant, we determined that [cit3–] was 0.025 M and [Hcit2–] was 2.5×10-5 K.
\[\dfrac{two.5 \times 10^{-v}}{0.025} \times 100 = 0.ane\%\]
If 0.i% is in the course Hcit2–, then 99.nine% is in the form cit3– and it simply makes it.
The other part of the problem was to calculate the concentration of all species in solution at the second equivalence signal. The outset thing we ought to do is compile a list of what all the species are and so nosotros know what we have to summate. In doing this, nosotros can ignore any spectator ions such as sodium. That ways there are vi species whose concentration we need to calculate.
H3cit Htwocit– Hcit2– cit3– H3O+ OH–
Since nosotros already calculated the pH of this solution (v.58), we can readily summate the concentration of H3O+ and OH–.
[H3O+] = 2.63×ten-vi
[OH–] = 3.80×10-9
We also said that to a get-go approximation it was "all" Hcit2– (0.0030 mol). With 30 ml of initial solution and lx ml of titrant, we accept a total volume of 90 ml.
[Hcit2–] = 0.0030 mol/0.090 L = 0.033 M
Since nosotros at present take [HthreeO+] and [Hcittwo–], nosotros can apply the appropriate Ka expressions to calculate the three other citrate species.
Utilise the Thoua3 expression to summate [citiii–]:
\[\mathrm{K_{a3} =\dfrac{[cit^{three-}][H_3O^+]}{[Hcit^{2-}]}=\dfrac{[cit^{3-}](ii.63\times10^{-6})}{0.033} = 4.02\times10^{-seven}}\]
\[\mathrm{[cit^{3-}] = 0.005\: M}\]
Use the Ka2 expression to summate [H2cit–]:
\[\mathrm{K_{a2} =\dfrac{[Hcit^{two-}][H_3O^+]}{[H_2cit^{2-}]}=\dfrac{(0.033)(2.63\times10^{-6})}{[Hcit^{ii-}]} = 1.733\times10^{-v}}\]
\[\mathrm{[H_2cit^–] = 0.005\: M}\]
Using the value of H2cit– that was just calculated, we tin substitute this into the Ka1 expression and calculate the concentration of H3cit.
\[\mathrm{K_{a1} =\dfrac{[H_2cit^{two-}][H_3O^+]}{[H_3cit]} =\dfrac{(0.005)(2.63\times10^{-6})}{[H_3cit]} = 7.45\times10^{-4}}\]
\[\mathrm{[H_3cit] = 1.77 \times x^{-five}}\]
One last set of things to examine are the calculated values for [H2cit–], [Hcitii–] and [citiii–].
[Hcit2–] = 0.033 Grand
[Hiicit–] = 0.005 M
[cit3–] = 0.005 M
What we demand to appreciate is that there is a problem with these numbers. We started this adding by challenge that "all" of the fabric was in the form of Hcitii–. What these calculations show is that there are appreciable amounts of H2cit– and cit3– in solution. If these two values were accurate, it would mean that the concentration of Hcit2– could only be 0.023 M.
Why does this happen? It has to practise with how close the pKa values are for citric acrid. The approximation that we could examine this every bit a stepwise manner, where we proceeded from ane reaction to the other and that intermediates were overwhelmingly the predominant form at the equivalence points, breaks downwards in this case because of how shut the pKa values are. The interesting function of this is that the pH of the solution would be 5.58, and that nosotros will get exactly equivalent concentrations of Htwocit– and cit3–, although they volition not exist exactly 0.005 Thousand. In reality, I do not think we would ever endeavour to summate the exact amount of each of these species at a pH like this, although afterwards in the grade we are going to come back to the citric acid state of affairs and meet a style to calculate the exact concentration of species present provided we know the pH.
One thing to keep in mind is that often times we do non use equilibrium calculations to arrive at exact values of substances. For one thing, concentrations are an approximation of activities and this may not ever be a expert one. For another, nosotros ofttimes use equilibrium calculations to provide ballpark values to let us know whether a particular process we may be considering is even feasible. In this instance, for case, these values show that we could never use the second equivalence point in a citric acrid titration for measurement purposes. This does not hateful that citric acrid cannot be used as a buffer, because it oft is. However, if we prepare a citric acid buffer (or any buffer), nosotros practise non rely on calculated amounts to ensure that the pH is where we want it, nosotros use a pH meter to monitor the buffer and use pocket-size amounts of a stiff acrid or strong base to accommodate the pH to the value we desire.
Titration Of H3po4 With Naoh,
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