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A buffer solution is prepared by adding \(0.125 \mathrm{~mol}\) ammonium chloride to 500. mL of 0.500 -M aqueous ammonia. Calculate the pH of the buffer. If 0.0100 mol HCl gas is bubbled into 500. mL buffer and all of the gas dissolves, calculate the new \(\mathrm{pH}\) of the solution.

Short Answer

Expert verified
Initial pH = 9.551; After HCl, pH = 9.810.

Step by step solution

01

Identify Components and Concentrations

Begin by identifying the components of the buffer solution. The buffer consists of aqueous ammonia (NH鈧), a weak base, and ammonium chloride (NH鈧凜l), its conjugate acid. We have 0.125 mol of NH鈧凜l added to 500 mL of 0.500 M NH鈧. Convert the volume to liters to make the calculations easier: 500 mL = 0.500 L.
02

Calculate Initial Concentrations

Calculate the initial concentration of NH鈧勨伜 ions from NH鈧凜l. Since 0.125 mol of NH鈧凜l is added to 0.5 L of solution, the concentration of NH鈧勨伜 is \(\frac{0.125}{0.5} = 0.25\) M. The initial concentration of NH鈧 is given as 0.500 M.
03

Use the Henderson-Hasselbalch Equation

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: \[pH = pK_a + \log \left(\frac{[\text{base}]}{[\text{acid}]}\right)\]Here, the base is NH鈧 and the acid is NH鈧勨伜. The \(pK_a\) of NH鈧勨伜 is 9.25. Substitute the concentrations into the equation:\[pH = 9.25 + \log \left(\frac{0.500}{0.250}\right)\]This simplifies to \(pH = 9.25 + 0.301 = 9.551\).
04

Calculate Moles of HCl Added

When HCl gas is bubbled into the solution, it reacts with the NH鈧 in the buffer. Calculate the number of moles of HCl added: 0.0100 mol.
05

Adjust Concentrations After Reaction with HCl

HCl will react with NH鈧 to form NH鈧勨伜. Thus, the moles of NH鈧 decrease by 0.0100 mol, and the moles of NH鈧勨伜 increase by 0.0100 mol. Adjust the concentrations:- New NH鈧 moles: 0.500 mol - 0.010 mol = 0.490 mol- New NH鈧勨伜 moles: 0.125 mol + 0.010 mol = 0.135 mol- Divide each by the volume (0.5 L) to find new concentrations: - [NH鈧僝 = \(\frac{0.490}{0.5} = 0.980\) M - [NH鈧勨伜] = \(\frac{0.135}{0.5} = 0.270\) M
06

Recalculate pH Using Updated Concentrations

Use the Henderson-Hasselbalch equation again with the updated concentrations:\[pH = 9.25 + \log \left(\frac{0.980}{0.270}\right)\]This simplifies to:\(pH = 9.25 + \log(3.63)\)\(pH = 9.25 + 0.560\)\(pH \approx 9.810\).

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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 fundamental tool in chemistry for calculating the pH of buffer solutions. It provides a clear relationship between the pH, the pK鈧 of the acid, and the concentration ratio of the base and its conjugate acid.

This equation is expressed as: \[ pH = pK_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \]This expression allows us to predict how changes in the concentrations of the components of the buffer affect the pH.
  • The \( pK_a \) is the negative logarithm of the acid dissociation constant, a measure of acid strength.
  • The fraction \( \frac{[\text{base}]}{[\text{acid}]} \) represents the relative concentrations of the base and the acid in the buffer solution.
This equation helps understand buffer capacity, which is how well a buffer can resist changes in pH. The closer the \( pH \) to the \( pK_a \), the greater the capacity of the buffer.
weak base
A weak base is a base that does not fully ionize in water. This means it does not completely convert into its constituent ions. Instead, weak bases exist in a dynamic equilibrium between the undissociated base and the ions.

In the provided exercise, we see ammonia (\( NH_3 \) ) functioning as a weak base.
  • A key property of weak bases is their partial ionization in water, forming hydroxide ions (\( OH^- \) ) and their conjugate acids.
  • The equilibrium constant for this process is termed the base ionization constant (\( K_b \) ).
For ammonia, when dissolved in water, the reaction is:\[ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \] This equilibrium is crucial for determining its behavior in buffer solutions. By understanding and using these principal properties, chemists can predict and control the pH levels in various chemical processes.
acid-base equilibria
Acid-base equilibria describe the balance between acids and bases in a solution, which is crucial for understanding the chemical nature of reactions. In the context of chemistry, it is all about how acid and base strengths influence reactions.

In any acidic or basic solution, there is a dynamic equilibrium between the proton donor (acid) and proton acceptor (base).
  • Equilibrium implies that the forward and reverse reactions occur at the same rate.
  • The concentrations of acids, bases, and their ions determine the exact position of the equilibrium.
Understanding this concept is essential when working with buffer solutions which involve the interplay of weak acids or bases and their conjugate salts to resist pH changes.
In the buffer solution of the exercise, the equilibrium involves:\[ NH_3 + H^+ \rightleftharpoons NH_4^+ \]It's important because it illustrates how adding acids or bases affects the equilibrium, and subsequently the pH. Having insight into acid-base equilibria allows chemists to control pH in industrial processes and biological systems.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) change when \(10.0 \mathrm{~mL}\) of \(0.100-\mathrm{M}\) \(\mathrm{NaOH}\) is added to \(90.0 \mathrm{~mL}\) pure water, and compare the \(\mathrm{pH}\) change with that when the same amount of \(\mathrm{NaOH}\) solution is added to \(90.0 \mathrm{~mL}\) of a buffer consisting of \(1.00-\mathrm{M} \mathrm{NH}_{3}\) and \(1.00-\mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\). Assume that the vol- umes are additive. \(K_{\mathrm{b}}\) of \(\mathrm{NH}_{3}=1.8 \times 10^{-5}\)

One liter \((1.0 \mathrm{~L})\) of distilled water has an initial \(\mathrm{pH}\) of 5.6 because it is in equilibrium with carbon dioxide in the atmosphere. One drop of concentrated hydrochloric acid, \(12-\mathrm{M} \mathrm{HCl}\), is added to the distilled water. Calculate the \(\mathrm{pH}\) of the resulting solution. 20 drops \(=1.0 \mathrm{~mL}\).

Predict what effect each would have on this equilibrium: $$ \mathrm{PbCl}_{2}(\mathrm{~s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) $$ (a) Addition of lead(II) nitrate solution (b) Addition of silver nitrate solution (c) Addition of \(\mathrm{NaCl}\) solution

The grid has six lettered boxes, each of which contains an item that may be used to answer the questions that follow. Items may be used more than once and there may be more than one correct item in response to a question. $$ \begin{array}{|l|l|l|} \hline \mathbf{A} & \mathbf{B} & \mathbf{C} \\ {\left[\mathrm{Cr}(\mathrm{OH})_{4}\right]^{-}} & Q>K_{\mathrm{sp}} & \mathrm{NaOH} \\ \hline \mathbf{D} & \mathbf{E} & \mathbf{F} \\ \text { Decreasing } \mathrm{pH} & \mathrm{pH}=\mathrm{p} K_{\mathrm{a}} & \begin{array}{l} \text { Sufficient } \mathrm{KCl} \\ \text { added to } \mathrm{CuCl}(\mathrm{aq}) \end{array} \\ \hline \end{array} $$ Place the letter(s) of the correct selection(s) on the appropriate line. (a) Could dissolve \(\mathrm{Zn}(\mathrm{OH})_{2}\) _________ (b) Could be used to prepare a buffer from \(\mathrm{HPO}_{4}^{2-}\) ___________ (c) Halfway to the equivalence point in the titration of a weak, monoprotic acid with strong base __________ (d) \(\mathrm{SO}_{2}\) -related atmospheric phenomenon _____________ (e) Species formed by a Lewis acid-base reaction ______________ (f) General condition required for precipitation to occur __________ (g) [conj. base] \(=[\) conj. acid \(]\) in a buffer ___________

Calculate the volume of \(0.150-\mathrm{M} \mathrm{HCl}\) required to titrate to the equivalence point for each of these samples. (a) \(25.0 \mathrm{~mL}\) of \(0.175-\mathrm{M} \mathrm{KOH}\) (b) \(15.0 \mathrm{~mL}\) of \(6.00-\mathrm{M} \mathrm{NH}_{3}\) (c) \(15.0 \mathrm{~mL}\) of propylamine, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\), which has a density of \(0.712 \mathrm{~g} / \mathrm{mL}\) (d) \(40.0 \mathrm{~mL}\) of \(0.0050-\mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\)

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