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A sample of a certain monoprotic weak acid was dissolved in water and titrated with 0.125 \(M\) NaOH, requiring \(16.00 \mathrm{mL}\) to reach the equivalence point. During the titration, the pH after adding \(2.00 \mathrm{mL}\) NaOH was \(6.912 .\) Calculate \(K_{\mathrm{a}}\) for the weak acid.

Short Answer

Expert verified
The Ka value for the weak acid is approximately \(1.74 \times 10^{-9}\).

Step by step solution

01

Determine the moles of OH- in the added NaOH

To calculate the moles of OH- in the added NaOH, we can use the given concentration and volume: Moles of OH- = \(concentration \times volume\) Moles of OH- = \(0.125 \frac{mol}{L} \times 0.002 L\) Moles of OH- = \(0.00025 mol\)
02

Calculate the moles of the weak acid present in the sample

Since we know that it required 16.00 mL of 0.125 M NaOH to reach the equivalence point, we can determine the moles of the weak acid in the sample: Moles of weak acid = moles of OH- (at equivalence point) Moles of weak acid = \(0.125 \frac{mol}{L} \times 0.016 L\) Moles of weak acid = \(0.002 mol\)
03

Calculate the initial concentrations of the weak acid and its conjugate base

After adding 2.00 mL of 0.125 M NaOH, the moles of the weak acid and its conjugate base will be: Moles of weak acid remaining = \(0.002 mol - 0.00025 mol = 0.00175 mol\) Moles of conjugate base formed = \(0.00025 mol\) Now, we can calculate the concentrations of the weak acid and its conjugate base: Initial concentration of the weak acid = \(\frac{0.00175 mol}{0.018 L} = 0.09722\frac{mol}{L}\) Initial concentration of the conjugate base = \(\frac{0.00025 mol}{0.018 L} = 0.01389\frac{mol}{L}\)
04

Use the provided pH and an equilibrium expression to determine the Ka value

From the given pH, we can determine the hydrogen ion concentration: pH = 6.912 \[H^+ = 10^{-pH} = 10^{-6.912} = 1.209 \times 10^{-7}\] Now, we can use the Ka expression to calculate Ka value: For a weak acid, HA, and its conjugate base, A-, the Ka expression is: \[K_a = \frac{[H^+][A^-]}{[HA]}\] Here, H+ is the hydrogen ion concentration, A- is the conjugate base concentration, and HA is the weak acid concentration. We can plug in the values: \[K_a = \frac{(1.209 \times 10^{-7})(0.01389)}{0.09722}\] \[K_a = 1.74 \times 10^{-9}\] So, the Ka value for the weak acid is approximately \(1.74 \times 10^{-9}\).

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Key Concepts

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

Weak Acid
A weak acid is an acid that only partially ionizes in water. This means it does not completely dissociate into its ions when dissolved. For example, if you dissolve 100 molecules of a weak acid in water, only a few of them will break apart and release hydrogen ions.
This is different from a strong acid that dissociates completely. Weak acids are essential in chemistry as they offer a buffer action and resist changes in pH. Typical weak acids include acetic acid, found in vinegar, and citric acid, present in citrus fruits.
  • Ionizes partially in solution
  • Selectively forms two types of ions
  • Has a measurable equilibrium with its ions
Understanding a weak acid's behavior is crucial while performing titration exercises, as it influences the pH of the solution throughout the experiment.
Equivalence Point
The equivalence point in a titration is a key concept. It is the point at which the amount of titrant added is just enough to completely neutralize the analyte solution.
In the case of an acid-base titration involving a weak acid and a strong base, the equivalence point marks the point when all the weak acid has reacted with the base to form water and a conjugate base.
For the given example, the equivalence point is reached with 16.00 mL of NaOH. At equivalence, the solution may not have a neutral pH due to the presence of the conjugate base in case of a weak acid:
  • Complete reaction between acid and base
  • Only product ions present
  • Important for calculating unknown concentrations
The pH at this point can provide crucial information and help calculate the acid's properties.
pH Calculation
Calculating the pH of a solution is a fundamental part of acid-base chemistry. pH is a measure of the hydrogen ion concentration in a solution.
The scale generally ranges from 0 to 14, with lower numbers being more acidic and higher numbers more basic. To find the pH, you can use the formula: \[ pH = -\log[H^+] \]
In the given example, after adding 2.00 mL of NaOH, the pH was measured to be 6.912. From this pH value, we can determine the concentration of hydrogen ions \([H^+]\):
\[H^+ = 10^{-6.912} = 1.209 \times 10^{-7}\] Understanding pH calculations is important because it helps us monitor how acidity changes during reactions, particularly in titration experiments.
Acid Dissociation Constant (Ka)
The acid dissociation constant, represented as \(K_a\), is a crucial value in chemistry. It measures the strength of an acid in solution and describes its ability to donate protons.
This constant is specific for each acid and defined by the equilibrium concentrations of the reactants and products in the ionization of the acid.
For a weak acid \( HA \) that dissociates into \( H^+ \) and \( A^- \), the expression is:\[K_a = \frac{[H^+][A^-]}{[HA]} \]In the given example, the \(K_a\) value was calculated using the concentrations at the specific titration point:\[K_a = \frac{(1.209 \times 10^{-7})(0.01389)}{0.09722}\]This gives a \(K_a\) of approximately \(1.74 \times 10^{-9}\).
  • Indicates how strongly an acid conducts electricity in a solution
  • Determines the degree of ionization for different acids
  • Helps predict the direction of the equilibrium reaction
The \(K_a\) value provides insights into both the nature of the acid itself and how it behaves in reactions.

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Most popular questions from this chapter

You have \(75.0 \mathrm{mL}\) of \(0.10 \mathrm{M}\) HA. After adding \(30.0 \mathrm{mL}\) of \(0.10 M \mathrm{NaOH},\) the \(\mathrm{pH}\) is \(5.50 .\) What is the \(K_{\mathrm{a}}\) value of \(\mathrm{HA} ?\)

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 \(0.200 M\) HNO \(_{3}\). Calculate the pH of the resulting solution after the following volumes of \(\mathrm{HNO}_{3}\) have been added. a. \(0.0 \mathrm{mL}\) b. \(20.0 \mathrm{mL}\) c. \(25.0 \mathrm{mL}\) d. \(40.0 \mathrm{mL}\) e. \(50.0 \mathrm{mL}\) f. \(100.0 \mathrm{mL}\)

A 0.210 -g sample of an acid (molar mass \(=192 \mathrm{g} / \mathrm{mol}\) ) is titrated with \(30.5 \mathrm{mL}\) of \(0.108 \mathrm{M} \mathrm{NaOH}\) to a phenolphthalein end point. Is the acid monoprotic, diprotic, or triprotic?

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 \(100.0 \mathrm{mL}\) NaOH added, what volume of NaOH added corresponds to the second equivalence point? b. For the following volumes of NaOH added, list the major species present after the OH \(^{-}\) reacts completely. i. \(0 \mathrm{mL} \mathrm{NaOH}\) added ii. between 0 and \(100.0 \mathrm{mL}\) NaOH added iii. \(100.0 \mathrm{mL}\) NaOH added iv. between 100.0 and \(200.0 \mathrm{mL}\) NaOH added v. \(200.0 \mathrm{mL} \mathrm{NaOH}\) added vi. after \(200.0 \mathrm{mL}\) NaOH added c. If the \(\mathrm{pH}\) at \(50.0 \mathrm{mL}\) 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.

A student dissolves 0.0100 mole of an unknown weak base in \(100.0 \mathrm{mL}\) water and titrates the solution with \(0.100 \mathrm{M} \mathrm{HNO}_{3}\) After \(40.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}\) was added, the \(\mathrm{pH}\) of the resulting solution was \(8.00 .\) Calculate the \(K_{\mathrm{b}}\) value for the weak base.

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