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Chapter 15: Problem 57

Consider the titration of \(40.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HClO}_{4}\) by \(0.100 M \mathrm{KOH}\). Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of KOH have been added. a. \(0.0 \mathrm{~mL}\) b. \(10.0 \mathrm{~mL}\) c. \(40.0 \mathrm{~mL}\) d. \(80.0 \mathrm{~mL}\) e. \(100.0 \mathrm{~mL}\)

Short Answer

Expert verified
The pH values at different stages of the titration process are: a. \(pH \approx 0.70\) b. \(pH \approx 1.30\) c. \(pH = 7\) d. \(pH \approx 12.69\) e. \(pH \approx 13.09\)

Step by step solution

01

Calculate the moles of the initial acid and base

Given the initial volume and concentration of the acid and the concentration of the base, we can determine the initial moles of HClO4 and KOH. \(Initial \ moles \ of \ HClO_4 \ (n_{HClO4}) = volume \times concentration = 40.0\mathrm{~mL} \times 0.200\mathrm{~M} \) \(Initial \ moles \ of \ KOH \ (n_{KOH}) = Added \ volume \times concentration \) Now, we can use these values to determine the pH at different stages of the titration process.
02

a. pH with 0.0 mL of KOH added

Since no KOH has been added yet, the pH is only determined by the initial HClO4 concentration. \(pH = -\log[H^+] = -\log(0.200\mathrm{~M})\)
03

b. pH with 10.0 mL of KOH added

First, calculate the moles of KOH added and subtract the number of moles of acid reacting with it. \(n_{KOH} = 10.0\mathrm{~mL} \times 0.100\mathrm{~M}\) \(n'_{HClO4} = n_{HClO4} - n_{KOH}\) Now, calculate the new concentration of HClO4 and then the pH. \(New \ concentration \ of \ HClO4 = \frac{n'_{HClO4}}{Initial \ volume \ + \ Added \ volume}\) \(pH = -\log[H^+]\)
04

c. pH with 40.0 mL of KOH added

This is the equivalence point, where all of the acid has been neutralized by the added base. The moles of the remaining solution are equal to the moles of the resulting salt, KClO4. Since KClO4 is a neutral salt, pH at this point will be equal to 7.
05

d. pH with 80.0 mL of KOH added

Calculate the excess KOH added and its new concentration in the solution. \(n_{KOH} = 80.0\mathrm{~mL} \times 0.100\mathrm{~M}\) \(Excess \ KOH = n_{KOH} - n_{HClO4}\) \(New \ concentration \ of \ OH^- = \frac{Excess \ KOH}{Initial \ volume \ + \ Added \ volume}\) Now, calculate the pOH using the new hydroxide ion concentration and then the pH. \(pOH = -\log[OH^-]\) \(pH = 14 - pOH\)
06

e. pH with 100.0 mL of KOH added

Repeat the step for 80.0 mL of KOH added, but replace the added volume with 100.0 mL. \(n_{KOH} = 100.0\mathrm{~mL} \times 0.100\mathrm{~M}\) \(Excess \ KOH = n_{KOH} - n_{HClO4}\) \(New \ concentration \ of \ OH^- = \frac{Excess \ KOH}{Initial \ volume \ + \ Added \ volume}\) \(pOH = -\log[OH^-]\) \(pH = 14 - pOH\)

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Key Concepts

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

Acid-Base Titration
Acid-base titration is a method used in chemistry to determine the concentration of an acid or a base in a solution. This process involves the gradual addition of one solution, known as the titrant, to another solution, which contains the substance of unknown concentration, with the aid of an indicator or a pH meter to detect the point of neutralization. This point, called the equivalence point, is reached when the number of moles of acid equals the number of moles of base, resulting in a neutral solution with pH 7 for strong acid and strong base reactions.

During titration, various pH levels are observed, depending on the amount of titrant added. If we were to use the example from the exercise with hydrochloric acid (HClO_4) and potassium hydroxide (KOH), the pH would change with every addition of KOH to the acidic solution of HClO_4. Initially, the solution is highly acidic, then gradually becomes less acidic, and at the equivalence point, it is neutral. Beyond that point, the solution turns basic as excess KOH is present.
Neutralization Reaction
In the context of acid-base chemistry, a neutralization reaction occurs when an acid and a base react to form water and a salt. This process effectively 'neutralizes' the acid and base, creating a solution with a pH level that depends on the strengths of the original reactants. Strong acids and strong bases will neutralize to produce a neutral pH (7) at the equivalence point, as seen in the reaction between HClO_4 and KOH.

Equivalence Point

The equivalence point of the titration is crucial as it signifies the exact point where the amount of acid is stoichiometrically equal to the amount of base in solution. By calculating the moles of HClO_4 and KOH initially present and the changes in concentration due to titration, as shown in the exercise steps, one can determine when the neutralization is complete. Precise knowledge of this point is important for understanding the reaction and for calculating concentrations after neutralization.
pH and pOH Calculations
The pH scale is a measure of the acidity or basicity of an aqueous solution, ranging from 0 to 14, with 7 being neutral. The pH is calculated as the negative logarithm of the hydrogen ion concentration (-[OH^-]). Conversely, the pOH represents the basicity and is calculated as the negative logarithm of the hydroxide ion concentration (-[OH^-]). The relationship between pH and pOH can be described by the equation pH + pOH = 14, which holds for aqueous solutions at 25°C.

Calculating pH and pOH

Using the provided exercise as an example, to find the pH at different titration stages, you first need to calculate the hydrogen or hydroxide ion concentrations. When acidic HClO_4 is present in excess, you calculate the pH directly from the hydrogen ion concentration. When basic KOH is in excess, you determine the pOH from the hydroxide ion concentration and then convert it to pH using the relation pH = 14 - pOH. These calculations reflect the changing acidic or basic nature of the solution during the titration process.

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Most popular questions from this chapter

When a diprotic acid, \(\mathrm{H}_{2} \mathrm{~A}\), is titrated with \(\mathrm{NaOH}\), the protons on the diprotic acid are generally removed one at a time, resulting in a pH curve that has the following generic shape: a. Notice that the plot has essentially two titration curves. If the first equivalence point occurs at \(100.0 \mathrm{~mL} \mathrm{NaOH}\) added, what volume of \(\mathrm{NaOH}\) added corresponds to the second equivalence point? b. For the following volumes of \(\mathrm{NaOH}\) added, list the major species present after the \(\mathrm{OH}^{-}\) reacts completely. i. \(0 \mathrm{~mL}\) NaOH added ii. between 0 and \(100.0 \mathrm{~mL}\) NaOH added iii. \(100.0 \mathrm{~mL} \mathrm{NaOH}\) added iv. between \(100.0\) and \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added v. \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added vi. after \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added c. If the \(\mathrm{pH}\) at \(50.0 \mathrm{~mL} \mathrm{NaOH}\) added is \(4.0\) and the \(\mathrm{pH}\) at \(150.0 \mathrm{~mL} \mathrm{NaOH}\) added is \(8.0\), determine the values \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) for the diprotic acid.

Carbonate buffers are important in regulating the \(\mathrm{pH}\) of blood at \(7.40\). If the carbonic acid concentration in a sample of blood is \(0.0012 M\), determine the bicarbonate ion concentration required to buffer the \(\mathrm{pH}\) of blood at \(\mathrm{pH}=7.40\). \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=4.3 \times 10^{-7}\)

Calculate the \(\mathrm{pH}\) of a solution prepared by mixing \(250 . \mathrm{mL}\) of \(0.174 m\) aqueous \(\mathrm{HF}\) (density \(=1.10 \mathrm{~g} / \mathrm{mL}\) ) with \(38.7 \mathrm{~g}\) of an aqueous solution that is \(1.50 \% \mathrm{NaOH}\) by mass (density \(=\) \(1.02 \mathrm{~g} / \mathrm{mL}) \cdot\left(K_{\mathrm{a}}\right.\) for \(\left.\mathrm{HF}=7.2 \times 10^{-4} .\right)\)

A buffer solution is prepared by mixing \(75.0 \mathrm{~mL}\) of \(0.275 \mathrm{M}\) fluorobenzoic acid \(\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{2} \mathrm{~F}\right)\) with \(55.0 \mathrm{~mL}\) of \(0.472 \mathrm{M}\) so- dium fluorobenzoate. The \(\mathrm{p} K_{\mathrm{a}}\) of this weak acid is \(2.90\). What is the \(\mathrm{pH}\) of the buffer solution?

Consider \(1.0 \mathrm{~L}\) of a solution that is \(0.85 \mathrm{M} \mathrm{HOC}_{6} \mathrm{H}_{5}\) and \(0.80 \mathrm{M} \mathrm{NaOC}_{6} \mathrm{H}_{5} .\left(K_{\mathrm{a}}\right.\) for \(\left.\mathrm{HOC}_{6} \mathrm{H}_{5}=1.6 \times 10^{-10} \cdot\right)\) a. Calculate the \(\mathrm{pH}\) of this solution. b. Calculate the \(\mathrm{pH}\) after \(0.10 \mathrm{~mole}\) of \(\mathrm{HCl}\) has been added to the original solution. Assume no volume change on addition of \(\mathrm{HCl}\). c. Calculate the \(\mathrm{pH}\) after \(0.20 \mathrm{~mole}\) of \(\mathrm{NaOH}\) has been added to the original buffer solution. Assume no volume change on addition of \(\mathrm{NaOH}\).
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