Problem 51 - General Chemistry: Principles and Modern Applications

Question

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.

Step-by-Step Solution

Answer

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 1: 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: \(2.5\, mmol / 0.100 \, M = 25.00\, mL\).

Step 2: 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 \(\mathrm{pOH} = 14 - pH\).

Step 3: 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.

Step 4: 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 \(\mathrm{HCl}\).

Key Concepts

pOH calculation

In a strong acid-strong base titration, understanding how to calculate the \(\mathrm{pOH}\) at various points is crucial. Before starting the titration, at zero mL of \(\mathrm{HCl}\), the solution contains only \(\mathrm{NaOH}\), and the concentration of hydroxide ions \( [\mathrm{OH^-}]\) equals the solution's molarity, which is \(0.100 \, M\). Therefore, the \(\mathrm{pOH}\) is given by the formula: \[ \mathrm{pOH} = - \log [\mathrm{OH^-}] = - \log (0.100) = 1.0,\] indicating the solution is basic at this point.
As \(\mathrm{HCl}\) is added but before the equivalence point, \(\mathrm{NaOH}\) is gradually neutralized, decreasing \( [\mathrm{OH^-}]\). It's essential to recalculate: \( [\mathrm{OH^-}] = \frac{\text{remaining moles of } \mathrm{NaOH}}{\text{total volume in liters}}\). As \(\mathrm{HCl}\) reaches the equivalence point, all hydroxide ions are neutralized, leading to \(\mathrm{pOH} = 7\) because the solution is now almost pure water. After the equivalence point, excess \(\mathrm{HCl}\) causes \(\mathrm{pOH} = 14 - \mathrm{pH}\), where \(\mathrm{pOH}\) decreases indicating increasing acidity.

equivalence point

The equivalence point in a strong acid-strong base titration is the stage where the moles of acid added equals the moles of base in the solution. This occurs because, at this juncture, the acid has completely neutralized the base. For the titration of \(25.00 \, \mathrm{mL}\) of \(0.100 \, \mathrm{M} \, \mathrm{NaOH}\) with \(0.100 \, \mathrm{M} \, \mathrm{HCl}\):

• The moles of \(\mathrm{NaOH}\) initially in the solution are \(0.100 \, M \times 25.00 \, mL = 2.5 \, \mathrm{mmol}\).
• At equivalence, precisely \(2.5 \, \mathrm{mmol}\) of \(\mathrm{HCl}\) will have been added.
• The total volume at this point is \(25.00 \, mL + 25.00 \, mL = 50.00\, mL\).
• Since both \(\mathrm{NaOH}\) and \(\mathrm{HCl}\) are strong, they dissociate completely, leaving the solution as neutral water and salt.
• Thus, \(\mathrm{pH}\) and \(\mathrm{pOH}\) are both approximately 7.

Remember, the equivalence point is not always at \(pH = 7\) for all reactions, but it is for strong acid and strong base titrations since they form water.

titration curve

A titration curve visually represents the changes in pH or \(\mathrm{pOH}\) during a titration as titrant is added. For a strong acid-strong base titration:
The x-axis often tracks the volume of acid or base added, and the y-axis shows either pH or \(\mathrm{pOH}\). The initial section of the curve will show a basic \(\mathrm{pOH}\) for the initial \(\mathrm{NaOH}\), around 1.0. As \(\mathrm{HCl}\) is added, \(\mathrm{pOH}\) rises slightly until just before the equivalence point, where it rapidly approaches 7. This happens quickly because strong acids and bases react completely, leading to sudden changes.
After the equivalence point, any additional \(\mathrm{HCl}\) results in a steeper increase in \(\mathrm{pOH}\), reflecting the acidic nature. To create a pH curve from a \(\mathrm{pOH}\) curve, simply subtract each \(\mathrm{pOH}\) value from 14 to get the corresponding pH. The shape of the curve will dramatically shift from a gradual rise to a sharp surge at the equivalence point, vividly depicting the reaction completion.