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Chapter 17: Problem 51

For the titration of \(25.00 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) with \(0.100 \mathrm{M} \mathrm{HCl},\) calculate the \(\mathrm{pOH}\) at a few representative points in the titration, sketch the titration curve of pOH versus volume of titrant, and show that it has exactly the same form as Figure \(17-9 .\) Then, using this curve and the simplest method possible, sketch the titration curve of pH versus volume of titrant.

Short Answer

Expert verified
The volume of \(0.100\, M\, HCl\) required to reach the equivalence point with \(25.00\, mL\) of \(0.100\, M\, NaOH\) is \(25.00\, mL\). The \(\mathrm{pOH}\) at the start of the titration is 0; just before the equivalence point is between 0 and 7; at the equivalence point is 7; and after the equivalence point is over 7. The shape of the \(\mathrm{pOH}\) graph is a sharp increase in \(\mathrm{pOH}\) at around the equivalence point. The graph of pH versus volume of \(\mathrm{HCl}\) is a mirror image, decreasing sharply at the equivalence point.

Step by step solution

01

Calculate the volume of HCl required to reach the equivalence point

At the equivalence point, all the \(\mathrm{NaOH}\) will be neutralized by the \(\mathrm{HCl}\). For a strong acid-strong base titration, this occurs when the moles of acid equals the moles of base. Using the formula moles = volume x molarity gives that: \(0.100\, M \times 25.00\, mL = 2.5\, mmol\). Therefore, we will need an equal amount of moles of \(\mathrm{HCl}\), which is \(2.5\, mmol\). Dividing by the molarity of \(\mathrm{HCl}\) gives the needed volume: \(2.5\, mmol / 0.100 \, M = 25.00\, mL\).
02

Calculate the \(\mathrm{pOH}\) at different points of the titration

At the start (0 \(\mathrm{mL}\) of \(\mathrm{HCl}\)), the \(\mathrm{pOH}\) will just be the \(– log\ [\mathrm{OH^-}],\) where \([OH^-] = 0.100\, M = 1.000\); thus \(\mathrm{pOH} = - log(1) = 0\). Before the equivalence point, some but not all \(\mathrm{NaOH}\) has been neutralized by the \(\mathrm{HCl}\). However, as we are dealing with strong acid and base, their ion concentrations will be the same as their molarity. Again, we use the \(\mathrm{pOH} = – log[\mathrm{OH^-}]\) formula. At the equivalence point (25.00 mL of \(\mathrm{HCl}\)), all \(\mathrm{NaOH}\) has been neutralized and we are left with a solution of water and \(\mathrm{NaCl}\), neither of which affect the pH or \(\mathrm{pOH}\). Thus, \(\mathrm{pOH} = 7\). After the equivalence point, there is an excess of \(\mathrm{HCl}\), thus \(\mathrm{pOH} = 14 - pH\).
03

Plot the \(\mathrm{pOH}\) on a graph

For this step, the volume of \(\mathrm{HCl}\) is plotted on the x-axis and the \(\mathrm{pOH}\) on the y-axis. Data points should be calculated for various volumes of \(\mathrm{HCl}\) and then connected with a smooth curve.
04

Convert the \(\mathrm{pOH}\) graph to a pH graph

Finally, using the relationship between pH and \(\mathrm{pOH}\) in water (pH + \(\mathrm{pOH}\) = 14), the graph can be converted to show pH instead of \(\mathrm{pOH}\). This is as simple as subtracting each \(\mathrm{pOH}\) value from 14 to find the corresponding pH value, and then plotting these on the graph of pH versus volume of \(\mathrm{HCl}\).

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Most popular questions from this chapter

In 1922 Donald D. van Slyke ( J. Biol. Chem., 52, 525) defined a quantity known as the buffer index: \(\beta=\mathrm{d} C_{\mathrm{b}} / \mathrm{d}(\mathrm{pH}),\) where \(\mathrm{d} C_{\mathrm{b}}\) represents the increment of moles of strong base to one liter of the buffer. For the addition of a strong acid, he wrote \(\beta=-\mathrm{d} C_{\mathrm{a}} / \mathrm{d}(\mathrm{pH})\) By applying this idea to a monoprotic acid and its conjugate base, we can derive the following expression: \(\beta=2.303\left(\frac{K_{w}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+\frac{\mathrm{CK}_{\mathrm{a}}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left(\mathrm{K}_{\mathrm{a}}+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\right)^{2}}\right)\) where \(C\) is the total concentration of monoprotic acid and conjugate base. (a) Use the above expression to calculate the buffer index for the acetic acid buffer with a total acetic acid and acetate ion concentration of \(2.0 \times 10^{-2}\) and a \(\mathrm{pH}=5.0\) (b) Use the buffer index from part (a) and calculate the \(\mathrm{pH}\) of the buffer after the addition of of a strong acid. (Hint: Let \(\left.\mathrm{d} C_{\mathrm{a}} / \mathrm{d}(\mathrm{pH}) \approx \Delta C_{\mathrm{a}} / \Delta \mathrm{pH} .\right)\) (c) Make a plot of \(\beta\) versus \(\mathrm{pH}\) for a \(0.1 \mathrm{M}\) acetic acid buffer system. Locate the maximum buffer index as well as the minimum buffer indices.

The effect of adding \(0.001 \mathrm{mol} \mathrm{KOH}\) to 1.00 Lof a solution that is \(0.10 \mathrm{M} \mathrm{NH}_{3}-0.10 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) is to (a) raise the pH very slightly; (b) lower the pH very slightly; (c) raise the pH by several units; (d) lower the pH by several units.

If an indicator is to be used in an acid-base titration having an equivalence point in the pH range 8 to 10 , the indicator must (a) be a weak base; (b) have \(K_{\mathrm{a}}=1 \times 10^{-9} ;(\mathrm{c})\) ionize in two steps; (d) be added to the solution only after the solution has become alkaline.

Consider a solution containing two weak monoprotic acids with dissociation constants \(K_{\mathrm{HA}}\) and \(K_{\mathrm{HB}}\). Find the charge balance equation for this system, and use it to derive an expression that gives the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) as a function of the concentrations of \(\mathrm{HA}\) and HB and the various constants.

Calculate the \(\mathrm{pH}\) at the points in the titration of \(25.00 \mathrm{mL}\) of \(0.132 \mathrm{M} \mathrm{HNO}_{2}\) when (a) \(10.00 \mathrm{mL}\) and (b) 20.00 mL of 0.116 M \(\mathrm{HNO}_{2}\) when \((\mathrm{a}) 10.00 \mathrm{mL}\) and (b) \(20.00 \mathrm{mL}\) of \(0.116 \mathrm{M}\) NaOH have been added. For \(\mathrm{HNO}_{2}, K_{\mathrm{a}}=7.2 \times 10^{-4}\) \(\mathrm{HNO}_{2}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NO}_{2}^{-}\)
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