Acids, Bases, and Solution Chemistry – Step-by-Step Guidance

Q1. What is the conjugate base of HPO42−?

Background

Topic: Acid-Base Chemistry (Conjugate Pairs)

This question tests your understanding of conjugate acids and bases, specifically how to identify the conjugate base of a given species.

Key Terms:

• Conjugate base: The species formed when an acid loses a proton (H+).

• HPO42−: Hydrogen phosphate ion.

Step-by-Step Guidance

1. Recall that a conjugate base is formed by removing one H+ from the acid.

2. Write the equation for the loss of a proton from HPO42−:

   $\text{HPO}_4^{2-} \rightarrow \text{PO}_4^{3-} + \text{H}^+$

3. Compare the answer choices to see which matches the formula for the conjugate base.

Try solving on your own before revealing the answer!

Q2. Which of the following is NOT a conjugate acid/base pair?

Background

Topic: Acid-Base Chemistry (Conjugate Pairs)

This question tests your ability to recognize conjugate acid/base pairs, which differ by one proton (H+).

Key Terms:

• Conjugate acid/base pair: Two species that differ by a single proton.

• Examples: NH4+/NH3, H2O/H+

Step-by-Step Guidance

1. For each pair, check if one species can be formed from the other by adding or removing an H+.

2. Write the chemical equations for each pair to see if they differ by a single proton.

3. Identify the pair that does NOT fit this definition.

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Q3. Calculate the pH of a solution that contains $7.8 \times 10^{-4}$ M OH−.

Background

Topic: Solution Chemistry (pH and pOH Calculations)

This question tests your ability to calculate pH from the concentration of hydroxide ions (OH−).

Key Formulas:

• $\text{pOH} = -\log[\text{OH}^-]$

• $\text{pH} + \text{pOH} = 14$ (at 25°C)

Step-by-Step Guidance

1. Calculate the pOH using the formula:

   $\text{pOH} = -\log(7.8 \times 10^{-4})$

2. Use the relationship between pH and pOH:

   $\text{pH} = 14 - \text{pOH}$

3. Set up the calculation for pH, but stop before plugging in the final value.

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Q4. Find the percent ionization of a 0.337 M HF solution. The Ka for HF is $3.5 \times 10^{-4}$.

Background

Topic: Weak Acid Equilibrium (Percent Ionization)

This question tests your ability to calculate the percent ionization of a weak acid using its concentration and acid dissociation constant (Ka).

Key Terms and Formulas:

• Percent ionization: $\frac{[\text{H}^+]}{[\text{HF}]_\text{initial}} \times 100\%$

• Ka expression for HF: $\text{Ka} = \frac{[\text{H}^+][\text{F}^-]}{[\text{HF}]}$

Step-by-Step Guidance

1. Set up the equilibrium expression for HF dissociation:

   $\text{HF} \rightleftharpoons \text{H}^+ + \text{F}^-$

2. Let $x$ be the amount of HF that ionizes. Write the Ka expression:

   $\text{Ka} = \frac{x^2}{0.337 - x}$

3. Assume $x$ is small compared to 0.337, so $0.337 - x \approx 0.337$.

4. Solve for $x$ (concentration of H+), but stop before calculating the final value.

Try solving on your own before revealing the answer!