Question

Consider the following six beakers. All have \(100 \mathrm{~mL}\) of aqueous \(0.1 \mathrm{M}\) solutions of the following compounds: beaker \(\mathrm{A}\) has \(\mathrm{HI}\) beaker \(\mathrm{B}\) has \(\mathrm{HNO}_{2}\) beaker \(\mathrm{C}\) has \(\mathrm{NaOH}\) beaker \(\mathrm{D}\) has \(\mathrm{Ba}(\mathrm{OH})_{2}\) beaker \(\mathrm{E}\) has \(\mathrm{NH}_{4} \mathrm{Cl}\) beaker \(\mathrm{F}\) has \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) Answer the questions below, using LT (for is less than), GT (for is greater than), \(\mathrm{EQ}\) (for is equal to), or MI (for more information required). (a) The \(\mathrm{pH}\) in beaker \(\mathrm{A}\) _____ the \(\mathrm{pH}\) in beaker \(\mathrm{B}\). (b) The \(\mathrm{pH}\) in beaker \(\mathrm{C}\) _____ the \(\mathrm{pH}\) in beaker \(\mathrm{D}\). (c) The \% ionization in beaker A _____ the \(\%\) ionization in beaker \(\mathrm{C}\). (d) The \(\mathrm{pH}\) in beaker \(\mathrm{B}\) _____ the \(\mathrm{pH}\) in beaker \(\mathrm{E}\). (e) The \(\mathrm{pH}\) in beaker \(\mathrm{E}\) _____ the \(\mathrm{pH}\) in beaker \(\mathrm{F}\). (f) The \(\mathrm{pH}\) in beaker \(\mathrm{C}\) _____ the \(\mathrm{pH}\) in beaker \(\mathrm{F}\)

Short answer

Question: Compare the pH and ionization percentages of the given beakers. Answer: a) The pH in Beaker A is less than the pH in Beaker B. b) The pH in Beaker C is greater than the pH in Beaker D. c) The % ionization of Beaker A is 100%; more information is required for Beaker C. d) More information is required to compare the pH of Beaker B and Beaker E. e) The pH in Beaker E is less than the pH in Beaker F. f) The pH in Beaker C is greater than the pH in Beaker F.

Step-by-step solution

Calculate pH of Beaker A

HI is a strong acid, meaning it fully dissociates in water. So the concentration of H+ ions will be 0.1 M. Therefore, we can calculate the pH using the formula \(pH = -\log[H^{+}] = -\log(0.1) = 1\).

Calculate pH of Beaker B

HNO\(_{2}\) is a weak acid, meaning it doesn't fully dissociate in water. Thus, we cannot immediately find the pH. However, we know that the pH of a weak acid is always higher than that of a strong acid with the same concentration.

Compare pH values

Since the pH of Beaker A is 1 and the pH of Beaker B is higher than that, we can conclude that the pH in beaker A is LT (less than) the pH in beaker B. #b) Compare pH of Beaker C and Beaker D#

Calculate pH of Beaker C

NaOH is a strong base, meaning it fully dissociates in water to produce OH- ions. The concentration of OH- ions is 0.1 M. We can calculate the pOH using the formula \(pOH = -\log[OH^{-}] = -\log(0.1) = 1\). Then, pH = 14 - pOH = 14 - 1 = 13.

Calculate pH of Beaker D

Ba(OH)\(_{2}\) is a strong base that also fully dissociates in water. However, since each Ba(OH)\(_{2}\) produces 2 OH- ions, the concentration of OH- is 2 * 0.1 M = 0.2 M. We can calculate the pOH using the formula \(pOH = -\log[OH^{-}] = -\log(0.2)\). Then, pH = 14 - pOH.

Compare pH values

Since the pH of Beaker C is 13 and the pH of Beaker D is slightly less than that, we can conclude that the pH in beaker C is GT (greater than) the pH in beaker D. #c) Compare % ionization of Beaker A and Beaker C#

Calculate % ionization of Beaker A

Since HI is a strong acid and fully ionizes, the % ionization is 100%.

Calculate % ionization of Beaker C

As Beaker C contains a strong base (NaOH), this question is not directly applicable as it refers to the ionization of acids. Therefore, we can label it as MI (more information required). #d) Compare pH of Beaker B and Beaker E#

Calculate pH of Beaker E

NH\(_{4}\)Cl is a salt of a weak base (NH\(_{3}\)) and a strong acid (HCl). It will form NH\(_{4}^{+}\) and Cl\(^{-}\) ions in water. NH\(_{4}^{+}\) being a weak acid will slightly ionize producing H\(^{+}\) ions. Since Beaker B and Beaker E both have weak acidic solutions, it is challenging to directly compare their pH without additional calculations.

Compare pH values

Due to the complexity of the comparison, we need to label this as MI (more information required). #e) Compare pH of Beaker E and Beaker F#

Calculate pH of Beaker F

C\(_{2}\)H\(_{5}\)NH\(_{2}\) is a weak base. Thus, its pH will be higher than 7 (neutral pH).

Compare pH values

Since Beaker E has a weak acidic solution (pH < 7) and Beaker F has a weak basic solution (pH > 7), we can conclude that the pH in beaker E is LT (less than) the pH in beaker F. #f) Compare pH of Beaker C and Beaker F#

Compare pH values

Since Beaker F has a weak basic solution (pH > 7) and Beaker C has a strong basic solution (pH = 13), we can conclude that the pH in beaker C is GT (greater than) the pH in beaker F.

Key concepts

Acid-Base Reactions

Understanding acid-base reactions is essential for mastering the concept of pH and the behavior of different substances in water. Acid-base reactions involve the exchange of protons (H⁺ ions) between substances. Acids are proton donors, which means they tend to release H⁺ ions in water, while bases are proton acceptors, binding with H⁺ ions to form water or hydroxyl ions (OH^-).

In our example, HI in beaker A and HNO₂ in beaker B are acids that react with water to varying extents. HI is a strong acid, completely dissociating into H⁺ and I^- ions, while HNO₂, a weak acid, only partially dissociates. The extent of dissociation affects the resulting pH, which is a measure of the acidity or basicity of the solution. A lower pH indicates a higher concentration of H⁺ ions, making the solution more acidic.

pH Calculation

The pH is a scale used to specify the acidity or basicity of an aqueous solution. It is calculated as the negative logarithm to base 10 of the activity of hydrogen ions (H⁺). The formula for pH calculation is as follows:
\[pH = -\text{log}[H^{+}]\]
For strong acids or bases, this calculation is straightforward because they disassociate completely. For example, we calculated the pH of beaker A containing HI to be 1, based on the formula. However, for weak acids and bases, such as HNO₂ and C₂H₅NH₂, respectively, the calculation is more complex and often requires additional steps, such as the application of equilibrium constants (Ka or Kb) or the use of the Henderson-Hasselbalch equation for buffer solutions.

The concept of pOH can also be utilized, which is the negative logarithm of the hydroxide ion concentration. The relationship between pH and pOH is given by: \[pH + pOH = 14\]This relationship helps to easily calculate the pH from pOH for basic solutions, as we did for beaker C containing NaOH. The calculations become more complicated for polyprotic bases, like Ba(OH)₂ in beaker D, which can release more than one hydroxide ion per molecule.

Strong and Weak Electrolytes

Electrolytes are compounds that conduct electricity when dissolved in water because they disassociate into ions. The strength of an electrolyte is defined by its ability to disassociate in solution. Strong electrolytes, such as HI, NaOH, and Ba(OH)₂, disassociate almost completely, resulting in a high concentration of ions, which corresponds to 100% ionization for HI in beaker A and full dissociation of NaOH in beaker C.

On the other hand, weak electrolytes, like HNO₂ and NH₄Cl, only partially disassociate and, as a result, have a much lower degree of ionization. The weak base C₂H₅NH₂ in beaker F is another example of a weak electrolyte. The percentage of ionization can give an idea of the strength of the acid or base, and this partially explains why the pH in beaker B, which contains a weak acid, cannot be immediately compared to that of a strong acid without additional information, as the acid-base reaction is incomplete.