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Chapter 14: Problem 38

A buffered solution is made by adding \(50.0 \mathrm{g} \mathrm{NH}_{4} \mathrm{Cl}\) to 1.00 \(\mathrm{L}\) of a 0.75-M solution of \(\mathrm{NH}_{3}\). Calculate the pH of the final solution. (Assume no volume change.)

Short Answer

Expert verified
The pH of the final buffered solution is approximately 9.155, calculated using the Henderson-Hasselbalch equation with concentrations of NH鈧勨伜 and NH鈧� in the final solution.

Step by step solution

01

Calculate the concentration of NH鈧勨伜 and NH鈧�

First, we'll find the concentration of NH鈧勨伜 from the ammonium chloride added. We are given 50.0 g of NH鈧凜l, and we need to convert it to moles and then find the concentration in the 1.00 L solution. Molar mass of NH鈧凜l = 14.0067 (N) + 1.00784 (x4 Hydrogen) + 35.453 (Cl) = 53.491 g/mol Moles of NH鈧凜l = mass / molar mass = 50.0 g / 53.491 g/mol = 0.9346 mol Concentration of NH鈧勨伜 = moles / volume = 0.9346 mol / 1.00 L = 0.9346 M Next, we're given the initial concentration of NH鈧�, which is 0.75 M. As we're assuming there is no volume change when NH鈧凜l is added, this concentration remains the same in the final solution.
02

Use the Henderson-Hasselbalch equation

Now that we have the concentrations of NH鈧勨伜 and NH鈧�, we can use the Henderson-Hasselbalch equation to calculate the pH of the buffered solution. The equation is: pH = pKa + log ([A鈦籡/[HA]) For NH鈧�/NH鈧勨伜, the pKa value is from the dissociation constant of NH鈧勨伜, which is: Ka = [H鈧僌鈦篯[NH鈧僝 / [NH鈧勨伜] pKa = -log(Ka) For NH鈧勨伜, Ka = 5.6 x 10鈦宦光伆, so pKa = -log(5.6 x 10鈦宦光伆) 鈮� 9.25. Plugging all values into the Henderson-Hasselbalch equation, we find: pH = 9.25 + log ([NH鈧僝/[NH鈧勨伜]) pH = 9.25 + log (0.75 M / 0.9346 M)
03

Calculate the pH

Now we can calculate the pH: pH = 9.25 + log (0.802) pH 鈮� 9.25 + (-0.095) pH 鈮� 9.155 The pH of the final buffered solution is approximately 9.155.

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Key Concepts

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

Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a formula that relates the pH of a solution to the pKa of the acid and the ratio of the concentrations of the deprotonated form (base) to the protonated form (acid). Specifically, it is expressed as:\[\begin{equation}\text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\end{equation}\]

This equation is particularly useful for buffered solutions, which are designed to resist changes in pH when small amounts of acid or base are added. A buffer typically consists of a weak acid and its conjugate base or a weak base and its conjugate acid. In the context of the exercise, ammonia (NH鈧�) represents the weak base, and ammonium (NH鈧勨伜) represents its conjugate acid.

To apply this equation, you need the pKa value, which comes from the acid鈥檚 dissociation constant (Ka). You also need the concentrations of the base \(A^-\) and the acid \(HA\). By plugging these values into the equation, you can solve for the pH of the solution. Understanding and applying the Henderson-Hasselbalch equation are essential for predicting the behavior of buffered solutions in various chemical contexts.

Acid-Base Equilibrium
The concept of acid-base equilibrium plays a central role in understanding how the Henderson-Hasselbalch equation works. Acid-base equilibrium refers to the state where the rates of the forward and reverse reactions of acid and base dissociation are equal, resulting in stable concentrations of the species involved.

For a simple acid dissociation, the equilibrium is given by:

\[\begin{equation}\text{HA} \leftrightarrow \text{H}^+ + \text{A}^-,\end{equation}\]

where HA is the weak acid, H鈦� is the hydrogen ion, and A鈦� is the conjugate base. The equilibrium constant for this reaction is the acid dissociation constant, Ka, and reflects the strength of the acid. The pKa is the negative logarithm of Ka and is used in the Henderson-Hasselbalch equation.

In the case of the buffered solution from the exercise, NH鈧勨伜 (the acid) is in equilibrium with NH鈧� (the base) and H鈧僌鈦� (the hydronium ion). By carefully balancing the concentrations of NH鈧勨伜 and NH鈧�, the solution can minimize pH changes, making it an effective buffer.

Molar Mass Calculation
Molar mass calculation is a fundamental skill in chemistry which allows scientists and students to convert between the mass of a substance and the amount of substance in moles. The molar mass of a compound is determined by summing the atomic masses of each element present in the molecule, multiplied by the number of atoms of that element in the molecule.

Example Calculation from Exercise

In our given exercise with NH鈧凜l, we calculate the molar mass as follows:

\[\begin{equation}\text{Molar mass of NH}_4\text{Cl} = \text{Molar mass of N} + 4 \times \text{Molar mass of H} + \text{Molar mass of Cl},\end{equation}\]

This calculation lets us convert the given mass of NH鈧凜l into moles, which is crucial for subsequent calculations involving concentration. In real-world applications, knowing how to calculate molar mass is essential for preparing solutions and for stoichiometric calculations in chemical reactions.

Concentration of Solutions
The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent. The most common units of concentration are molarity (M), which is expressed as moles of solute per liter of solution. Understanding how to calculate and apply concentrations is critical for tasks such as mixing solutions and calculating pH values.

How to Calculate Concentration

The concentration is calculated using the formula:

\[\begin{equation}\text{Concentration (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\end{equation}\]

From our exercise, once the molar mass of NH鈧凜l is known, we can find the moles of NH鈧凜l, and then divide by the volume of the solution to get the concentration.

Applying these calculations to the buffer solution, you can determine the concentrations of both the weak base NH鈧� and the conjugate acid NH鈧勨伜, which are necessary inputs for the Henderson-Hasselbalch equation to calculate the pH. This process highlights the interconnectivity between the concepts of molar mass, concentration, and acid-base equilibrium in the study of chemistry.

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Most popular questions from this chapter

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.

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: $$\begin{aligned} &\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}(s)+2 \mathrm{OH}^{-}(a q)\\\&\text { Aspirin } \quad \frac{\text { Boil }}{10 \min } \underset{\text { Salicylate ion }}{\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{3}^{-}(a q)}+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \end{aligned}$$ A sample of aspirin with a mass of 1.427 g was boiled in \(50.00 \mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{NaOH} .\) After the solution was cooled, it took \(31.92 \mathrm{mL}\) of \(0.289 \mathrm{M}\) HCl to titrate the excess NaOH. Calculate the purity of the aspirin. What indicator should be used for this titration? Why?

Which of the following mixtures would result in a buffered solution when 1.0 L of each of the two solutions are mixed? a. \(0.2 M\) HNO and \(0.4 M \mathrm{NaNO}_{3}\) b. \(0.2 M \mathrm{HNO}_{3}\) and \(0.4 \mathrm{M} \mathrm{HF}\) c. \(0.2 \mathrm{M} \mathrm{HNO}_{3}\) and \(0.4 \mathrm{M} \mathrm{NaF}\) d. \(0.2 M\) HNO \(_{3}\) and \(0.4 M\) NaOH

Sketch two pH curves, one for the titration of a weak acid with a strong base and one for a strong acid with a strong base. How are they similar? How are they different? Account for the similarities and the differences.

Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intracellular fluids at \(\mathrm{pH}\) values generally between 7.1 and 7.2 a. What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) in intracellular fluid at \(\mathrm{pH}=7.15 ?\) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8}\) b. Why is a buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ineffective in buffering the \(\mathrm{pH}\) of intracellular fluid? \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=7.5 \times 10^{-3}\)
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