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Which of the following solutions is a buffer? (a) A solution made by mixing \(50 \mathrm{~mL}\), of \(0.200 \mathrm{M}\) formic acid \((\mathrm{HCOOH})\) and \(250 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KOH},(\mathbf{b})\) A solution made by \(\mathrm{mix}\) ing \(50 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) formic acid \((\mathrm{HCOOH})\) and \(25 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) nitric acid \(\left(\mathrm{HNO}_{3}\right),(\mathbf{c})\) A solution made by mixing \(50 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) potassium formate \((\mathrm{HCOOK})\) and \(25 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KNO}_{3},\) (d) A solution made by mixing \(50 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) formic acid \((\mathrm{HCOOH})\), and \(25 \mathrm{~mL}\). of \(0.200 \mathrm{MKOH}\).

Short Answer

Expert verified
Solution (d) is a buffer.

Step by step solution

01

Understand Buffer Solutions

A buffer solution is one that can resist changes in pH upon the addition of small amounts of an acid or a base. A buffer is typically made from either a weak acid and its conjugate base or a weak base and its conjugate acid. Identifying such pairs in the given solutions is key to determining if a buffer can be formed.
02

Analyze Solution (a)

Mixing 50 mL of 0.200 M formic acid with 250 mL of 0.200 M KOH results in an excess of KOH. This results in complete neutralization of formic acid, forming formate ions, but the presence of excess strong base means this is not a buffer solution.
03

Analyze Solution (b)

Solution (b) involves mixing formic acid with nitric acid, both of which are acids. There is no weak acid and conjugate base pair present, which means this solution can't act as a buffer.
04

Analyze Solution (c)

Solution (c) involves potassium formate (a salt of a weak acid, formic acid) and potassium nitrate. There is no conjugate acid-base pair, rendering this incapable of acting as a buffer solution despite containing the salt of a weak acid.
05

Analyze Solution (d)

Mixing formic acid with KOH allows for partial neutralization of the acid, producing formate ions (the conjugate base of formic acid). The presence of both the weak acid (HCOOH) and its conjugate base (HCOO-) allow this solution to resist changes in pH, making it a buffer solution.

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Key Concepts

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

pH resistance
Buffers have a special ability to resist pH changes, which makes them incredibly valuable in both laboratory and biological settings. This ability is due to the presence of a weak acid and its conjugate base. In simple terms, when you add a small amount of acid or base to a well-balanced buffer solution, the pH doesn't change much. This is because the components of the buffer react to neutralize the added substances.

For example, in the context of our original problem, a solution that partially neutralizes formic acid with potassium hydroxide (KOH) results in a mixture containing both the weak acid, formic acid, and its conjugate base, formate anion. This combination can absorb added H+ or OH- ions with minimal change in pH.

So, when a small amount of acid is added to the solution, the conjugate base picks up the added hydrogen ions (H+). Similarly, if a base is added, the weak acid can donate hydrogen ions to offset the increase in OH- ions. This balance is the secret to how buffers stabilize the pH of a solution.
weak acid and conjugate base
Buffer solutions derive their stabilizing power from a weak acid and its conjugate base. This pair works in harmony to keep pH levels steady. Let's break down what a weak acid and conjugate base are and how they function together.

A weak acid, like formic acid, only partially ionizes in water. This means it doesn't completely dissociate into ions in solution. When in solution, formic acid exists along with its conjugate base, the formate ion, created when the acid donates a proton (H+).

The conjugate base is crucial. It can accept hydrogen ions when the solution is too acidic. Conversely, if the solution is too basic, the weak acid can donate hydrogen ions. This creates a dynamic equilibrium that can effectively buffer the solution, preventing drastic changes in pH.

The original exercise highlights how solutions can promote buffering through the combination of weak acids and their conjugate bases. Specifically, in solution (d), mixing formic acid with a precise amount of KOH leads to the formation of both HCOOH and HCOO-, thus establishing a classic buffer pair.
neutralization reactions
Neutralization reactions are key in understanding how buffer solutions are formed. These reactions involve an acid and a base reacting to form water and a salt. In buffered solutions, partial neutralization occurs, ensuring both components of the buffer pair remain.

Consider solution (d) from the exercise. Formic acid reacts with KOH, a strong base, in a way that only part of the acid is neutralized. This reaction is selective and stops once the optimal balance of acid and conjugate base is formed. The equation for this can be represented as:
\[ \text{HCOOH} + \text{KOH} \rightarrow \text{HCOO}^- + \text{H}_2\text{O} + \text{K}^+ \]

Here, HCOOH (formic acid) reacts with KOH to produce water, along with potassium ions (K+), and the formate ion \( \text{HCOO}^- \), the conjugate base. However, because not all of the acid is neutralized, both the weak acid and its conjugate base exist together in the solution.

This reaction ensures that when additional acid or base is introduced, the buffer can mitigate pH changes effectively through further neutralization. This delicate balance is why buffer solutions are so powerful in resisting sudden pH shifts.

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Most popular questions from this chapter

Lead(II) carbonate, \(\mathrm{PbCO}_{3}\), is one of the components of the passivating layer that forms inside lead pipes. (a) If the \(K_{i p}\) for \(\mathrm{PbCO}_{3}\) is \(7.4 \times 10^{-14}\) what is the molarity of \(\mathrm{Pb}^{2+}\) in a saturated solution of lead(II) carbonate? (b) What is the concentration in ppb of \(\mathrm{Pb}^{2+}\) ions in a saturated solution? (c) Will the solubility of \(\mathrm{PbCO}_{3}\) increase or decrease as the \(\mathrm{pH}\) is lowered? (d) The EPA threshold for acceptable levels of lead ions in water is 15 ppb. Does a saturated solution of lead(II) carbonate produce a solution that exceeds the EPA limit?

A solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added dropwise to a solution that is \(0.010 \mathrm{M}\) in \(\mathrm{Ba}^{2+}(a q)\) and \(0.010 \mathrm{M}\) in \(\mathrm{Sr}^{2+}(a q)\). (a) What concentration of \(\mathrm{SO}_{4}^{2-}\) is necessary to begin precipitation? (Neglect volume changes. BaSO \(_{4}: K_{i p}=1.1 \times 10^{-10} ; \mathrm{SrSO}_{4}\) \(\left.K_{s p}=3.2 \times 10^{-7} .\right)(\mathbf{b})\) Which cation precipitates first? (c) What is the concentration of \(\mathrm{SO}_{4}^{2-}(a q)\) when the second cation begins to precipitate?

Which of the following solutions is a buffer? (a) \(0.20 \mathrm{M}\) formic acid (HCOOH), (b) \(0.20 \mathrm{M}\) formic acid \((\mathrm{HCOOH})\) and \(0.20 \mathrm{M}\) sodium formate (HCOONa), (c) \(0.20 \mathrm{M}\) nitric acid \(\left(\mathrm{HNO}_{3}\right)\) and \(0.20 \mathrm{M}\) sodium nitrate \(\left(\mathrm{NaNO}_{3}\right),\) (d) both b and \(\mathrm{c},(\mathbf{e})\) all of \(\mathrm{a}, \mathrm{b},\) and \(\mathrm{c}\)

Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\) and whose corresponding \(K_{\text {ip }}=6.8 \times 10^{-27}\). As discussed in the Chemistry and Life box on page 790 , fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F},\) whose \(K_{y p}=1.0 \times 10^{-60}\) (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

For each statement, indicate whether it is true or false. (a) The solubility of a slightly soluble salt can be expressed in units of moles per liter. (b) The solubility product of a slightly soluble salt is simply the square of the solubility. (c) The solubility of a slightly soluble salt is independent of the presence of a common ion. (d) The solubility product of a slightly soluble salt is independent of the presence of a common ion.

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