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Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HNO}_{3}\) with 0.100 \(\mathrm{M} \mathrm{NaOH} .\)

Short Answer

Expert verified
In this titration, 0.100 M HNO3 is reacted with 0.100 M NaOH. Before the equivalence point, the pH is determined by the concentration of HNO3. At the equivalence point, which occurs at 25.0 mL of NaOH, the pH will be 7 since NaNO3 is a neutral salt. After the equivalence point, the pH is determined by the excess NaOH. A titration curve can be plotted to observe the pH changes throughout the titration process.

Step by step solution

01

Understand the chemical reaction of the titration

In this titration, a strong acid (HNO3) is reacted with a strong base (NaOH). The reaction that occurs during titration is as follows: \[ \mathrm{HNO_3} + \mathrm{NaOH} \rightarrow \mathrm{NaNO_3} + \mathrm{H_2O} \]
02

Calculate initial moles of acid and base

Before starting the titration, we need to find the initial moles of HNO3 and NaOH. We can use the given volumes and concentrations for this calculation. Moles of HNO3 = Concentration × Volume Moles of HNO3 = 0.100 M × 25.0 mL × (1 L/1000 mL) = 0.00250 mol
03

Calculate the reaction progress and equivalence point

During the titration, the strong base (NaOH) is slowly added to the strong acid (HNO3). The moles of HNO3 and NaOH will react in a 1:1 ratio according to the balanced chemical equation. To find the equivalence point, which is when the moles of acid and base are equal, we need to calculate the volume of NaOH required for this to happen. Since both the acid and base have the same concentration (0.100 M), the volume of NaOH required for the equivalence point will be the same as the volume of HNO3, which is 25.0 mL.
04

Calculate the pH at different stages of titration

The pH of the solution will change as the titration progresses. We can break down the titration process into three main stages: before the equivalence point, at the equivalence point, and after the equivalence point. 1. Before the equivalence point: At this stage, the unreacted HNO3 will contribute to the pH of the solution. Since HNO3 is a strong acid, it dissociates completely in water. \[ \mathrm{[H^+]} = \mathrm{concentration\:of\:HNO_3} \] Using the pH formula, where pH = -log([H+]), we can calculate the pH of the solution. 2. At the equivalence point: The moles of HNO3 and NaOH are equal at this point, meaning all the HNO3 has been neutralized. The solution consists of NaNO3 and H2O. Since NaNO3 is a neutral salt and does not contribute to the pH, the pH at the equivalence point will be 7. 3. After the equivalence point: At this stage, there is an excess of NaOH, which determines the pH of the solution. We can calculate the moles of remaining NaOH and use the formula pH = 14 - pOH to find the final pH. To calculate pOH, use the formula pOH = -log([OH−]), where [OH−] is the concentration of the hydroxide ion in the solution.
05

Analyze the titration curve

Once we have calculated the pH at different stages of titration, we can plot a titration curve, which shows the changes in pH as the titration progresses. The curve will have a steep rise in pH around the equivalence point and gradually level off before and after the equivalence point, indicating the changes in pH as the strong acid (HNO3) is neutralized by the strong base (NaOH).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

pH Calculation
Understanding how to calculate the pH of a solution during a titration is essential. In acid-base titration, the pH varies depending on how many reactants remain and how much product is formed.
  • Before the equivalence point: At this stage, there is an excess of the strong acid from our example, HNO3. This means the pH will be determined by the concentration of the H+ ions, given that HNO3 is a strong acid and dissociates completely. To find the pH, use the formula:

    \[ pH = -\log([H^+]) \]

    where [H+] is the concentration of HNO3.
  • At the equivalence point: Here, equal amounts of the acid and the base have reacted. The resulting solution contains only products such as water and neutral salts, in this case, NaNO3. Since NaNO3 does not affect the pH, the solution is neutral with a pH of 7.
  • After the equivalence point: Now, the titration solution will contain an excess of the strong base, NaOH. The concentration of OH ions will determine the basic nature of the solution. To calculate the pH:

    First, find the pOH using \[ pOH = -\log([OH^-]) \]

    where[OH^-] is the concentration of the excess NaOH. Finally, calculate pH using:

    \[ pH = 14 - pOH \]

Equivalence Point
The equivalence point in an acid-base titration is when the amount of acid equals the amount of base added, resulting in complete neutralization. For a titration involving a strong acid and a strong base like HNO3 with NaOH, determining the equivalence point is straightforward since they react in a 1:1 molar ratio.
In our example, this means that the moles of HNO3 will equal the moles of NaOH at the equivalence point. Because the concentrations of HNO3 and NaOH are both 0.100 M, we can deduce that the volume of NaOH required to reach the equivalence point will also match the initial volume of HNO3, which is 25.0 mL.
Therefore, the total volume of solution at the equivalence point is 50.0 mL, and the chemical reaction is at complete balance. The result is a neutral solution with a pH of precisely 7, as only neutral products like NaNO3 and water remain. It's critical to remember that the equivalence point is not the same as the endpoint, which refers to the point where the indicator changes color.
Titration Curve
A titration curve is a graphical representation showing how the pH of a solution changes as a titrant is added. This graph is very useful for visualizing the titration process and for identifying key points such as the equivalence point.
In the case of strong acid-strong base titrations, like that of HNO3 with NaOH, the titration curve will have distinct features:
  • At the starting point, the pH is low because of the high concentration of the strong acid.
  • As the NaOH is added, you will see a slight, gradual rise in the pH. This represents the stage before reaching the equivalence point, where there is excess acid.
  • Once we near the equivalence point, the curve becomes much steeper, indicating a rapid increase in pH over a small addition of NaOH. This steep jump signifies the neutralization is close to completion.
  • After passing the equivalence point, the curve levels off again. This flatter region indicates the excess of base in the solution, leading to a higher pH.
The shape and shift of the curve offer valuable insight into the reaction dynamics. In particular, the steep vertical section around the equivalence point provides a clear indicator for determining the neutral point of the reaction.

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Most popular questions from this chapter

One method for determining the purity of aspirin \(\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right)\) is to hydrolyze it with NaOH solution and then to titrate the remaining NaOH. The reaction of aspirin with NaOH is as follows: \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}(s)+2 \mathrm{OH}^{-}(a q)\) $$ \mathrm{C}_{7} \mathrm{H}_{3} \mathrm{O}_{3}^{-}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ A sample of aspirin with a mass of 1.427 g was boiled in 50.00 \(\mathrm{mL}\) of 0.500\(M \mathrm{NaOH}\) . After the solution was cooled, it took 31.92 \(\mathrm{mL}\) of 0.289 \(\mathrm{M} \mathrm{HCl}\) to titrate the excess NaOH. Calculate the purity of the aspirin. What indicator should be used for this titration? Why?

What concentration of \(\mathrm{NH}_{4} \mathrm{Cl}\) is necessary to buffer a \(0.52-M\) \(\mathrm{NH}_{3}\) solution at \(\mathrm{pH}=9.00 ?\left(K_{\mathrm{b}} \text { for } \mathrm{NH}_{3}=1.8 \times 10^{-5} .\right)\)

Consider the titration of 100.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) \(\left(K_{\mathrm{b}}=3.0 \times 10^{-6}\right)\) by 0.200\(M \mathrm{HNO}_{3}\) . Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of \(\mathrm{HNO}_{3}\) have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 40.0 \mathrm{mL}} \\ {\text { b. } 20.0 \mathrm{mL}} & {\text { e. } 50.0 \mathrm{mL}} \\ {\text { c. } 25.0 \mathrm{mL}} & {\text { f. } 100.0 \mathrm{mL}}\end{array} $$

Mixing together solutions of acetic acid and sodium hydroxide can make a buffered solution. Explain. How does the amount of each solution added change the effectiveness of the buffer?

a. Calculate the \(\mathrm{pH}\) of a buffered solution that is 0.100 \(\mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5} )\) and 0.100 \(\mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}\) b. Calculate the pH after 20.0\(\%\) (by moles) of the benzoic acid is converted to benzoate anion by addition of a strong base. Use the dissociation equilibrium $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q) $$ to calculate the pH. c. Do the same as in part b, but use the following equilibrium to calculate the \(\mathrm{pH} :\) $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q) $$ d. Do your answers in parts \(b\) and \(c\) agree? Explain.

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