G12 C4 Acid Base Chemistry Short Question Answer

Chapter 4: Acid Base Chemistry

Short Question Answers

Question v: Mixing Acid Solutions

Question: Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the H$^+$ ion concentration in the mixture?

Answer:

First, we need to find the hydrogen ion concentration for each pH value. The formula relating pH and H$^+$ concentration is:
$$ pH = - \log_{10}[H^+] $$
Therefore, the concentration is:
$$ [H^+] = 10^{-pH} $$

• For pH 3: $[H^+]_1 = 10^{-3}$ M
• For pH 4: $[H^+]_2 = 10^{-4}$ M
• For pH 5: $[H^+]_3 = 10^{-5}$ M

Since equal volumes are mixed, we can assume each volume is $V$. The total moles of H$^+$ ions is the sum of the moles from each solution. Moles = Concentration × Volume.
Total moles of H$^+$ = $(10^{-3} \times V) + (10^{-4} \times V) + (10^{-5} \times V)$
$$ = V(10^{-3} + 10^{-4} + 10^{-5}) $$ $$ = V(0.001 + 0.0001 + 0.00001) $$ $$ = V(0.00111) $$

The total volume of the mixture is $V_{total} = V + V + V = 3V$.

The new concentration of H$^+$ ions in the mixture is:
$$ [H^+]_{mixture} = \frac{\text{Total moles of H}^+}{\text{Total volume}} $$ $$ [H^+]_{mixture} = \frac{V(0.00111)}{3V} = \frac{0.00111}{3} $$ $$ [H^+]_{mixture} = 0.00037 $$
The H$^+$ ion concentration in the mixture is $3.7 \times 10^{-4}$ M.

Question vi: Titration of Weak Base with Strong Acid

Question: A $20.0 \text{ cm}^3$ sample of $0.200 \text{ mol dm}^{-3} \text{ NH}_3(\text{aq})$ was titrated with $0.100 \text{ mol dm}^{-3} \text{ HCl}$. On the following axes, sketch how the pH changes during this titration. Mark clearly where the end point occurs.

Answer:

This is a titration of a weak base ($\text{NH}_3$) with a strong acid ($\text{HCl}$). The initial pH will be basic, but less than 14. As the strong acid is added, the pH will gradually decrease, forming a buffer region. The equivalence point will be acidic, as the products of the reaction ($\text{NH}_4\text{Cl}$) form an acidic solution. Finally, the pH will level out as the strong acid is in excess. The titration curve will look like the sketch below.

The equivalence point occurs when the moles of acid added equal the moles of the initial base.
Moles of $\text{NH}_3$ = Concentration × Volume
$$ = 0.200 \text{ mol dm}^{-3} \times 20.0 \text{ cm}^3 $$ $$ = 0.200 \times 0.0200 \text{ dm}^3 = 0.004 \text{ moles} $$

Volume of $\text{HCl}$ needed = $\frac{\text{Moles}}{\text{Concentration}}$
$$ = \frac{0.004 \text{ moles}}{0.100 \text{ mol dm}^{-3}} = 0.040 \text{ dm}^3 = 40.0 \text{ cm}^3 $$

The **end point** (or equivalence point) will be marked at **$40.0 \text{ cm}^3$ of added acid** and a pH below 7.

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3. Long Answer Questions

i. The Concept of pH

Question: Explain the concept of pH and its significance in acid-base chemistry. How does the pH scale relate to the concentration of hydrogen ions in a solution?

Answer:

The **pH** scale is a logarithmic scale used to specify the acidity or basicity of an aqueous solution. It's a way to express a wide range of hydrogen ion concentrations ($[H^+]$) in a more manageable and convenient way. The term pH stands for "potential of hydrogen" or "power of hydrogen."

The significance of pH in acid-base chemistry is fundamental. It provides a quick and universal measure of how acidic or basic a substance is, which in turn determines its chemical reactivity and biological compatibility. For example, many biological processes, like enzyme function in the human body, are highly sensitive to pH changes. Pure water has a pH of 7, which is considered **neutral**. Solutions with a pH below 7 are **acidic**, and those with a pH above 7 are **basic** (or alkaline).

The pH scale is directly related to the concentration of hydrogen ions ($[H^+]$). The relationship is defined by the following equation:
$$ pH = - \log_{10}[H^+] $$

This logarithmic relationship means that a change of one unit on the pH scale represents a tenfold change in the hydrogen ion concentration. For example, a solution with a pH of 4 has 10 times the hydrogen ion concentration of a solution with a pH of 5, and 100 times the concentration of a solution with a pH of 6.

ii. Comparing Titration Curves

Question: Compare and contrast the titration curves for a strong acid with a strong base, a weak acid with a strong base, a strong acid with a weak base, and a weak acid with a weak base.

Answer:

Titration curves are plots of pH versus the volume of titrant added, and their shapes reveal important information about the strength of the acid and base involved.

• Strong Acid-Strong Base: This curve starts with a very low pH (highly acidic) and ends with a very high pH (highly basic). The most prominent feature is a **very sharp and vertical pH change** around the equivalence point. The equivalence point for this type of titration is at **pH 7.0**, as the products (a salt and water) are neutral.
• Weak Acid-Strong Base: The curve starts at a higher pH than the strong acid-strong base curve. It features a **buffer region** (a section with a very gradual pH change) before the equivalence point, where the weak acid and its conjugate base are both present. The equivalence point is **greater than pH 7.0** due to the hydrolysis of the conjugate base. The pH change at the equivalence point is less sharp and shorter than in the strong acid-strong base titration.
• Strong Acid-Weak Base: This is essentially a mirror image of the weak acid-strong base curve. It starts at a lower pH than a weak acid and has a buffer region after the initial drop in pH. The equivalence point is **less than pH 7.0** due to the hydrolysis of the conjugate acid. The pH change at the equivalence point is also less sharp.
• Weak Acid-Weak Base: These titrations are generally not very useful for analytical purposes because they lack a sharp, clear equivalence point. The pH change around the equivalence point is very gradual and drawn out. There is no sharp "jump" in pH, making it difficult to pinpoint the equivalence point with a pH indicator. The equivalence point is typically near **pH 7.0**, but this depends on the relative strengths of the acid and base.

iii. Titration of Acetic Acid with Sodium Hydroxide

Question: A $50 \text{ cm}^3$ solution of $0.1 \text{ M}$ acetic acid is titrated with $0.1 \text{ M}$ sodium hydroxide. The $K_a$ of acetic acid at $25^\circ\text{C}$ is $1.8 \times 10^{-5}$. Calculate the pH of the solution at the equivalence point and explain the shape of the titration curve for this reaction.

Answer:

This is a titration of a **weak acid (acetic acid)** with a **strong base (sodium hydroxide)**.

Calculation of pH at the Equivalence Point

At the equivalence point, all of the acetic acid ($\text{CH}_3\text{COOH}$) has reacted with the sodium hydroxide ($\text{NaOH}$) to form sodium acetate ($\text{CH}_3\text{COONa}$) and water.
$$ \text{CH}_3\text{COOH}(\text{aq}) + \text{NaOH}(\text{aq}) \rightarrow \text{CH}_3\text{COONa}(\text{aq}) + \text{H}_2\text{O}(l) $$

First, we find the moles of the acid:
Moles of $\text{CH}_3\text{COOH} = 0.1 \text{ M} \times 0.050 \text{ dm}^3 = 0.005 \text{ moles}$

At the equivalence point, the moles of $\text{NaOH}$ added will also be $0.005 \text{ moles}$.
Volume of $\text{NaOH}$ added = $\frac{0.005 \text{ moles}}{0.1 \text{ M}} = 0.050 \text{ dm}^3 = 50 \text{ cm}^3$

The total volume of the solution at the equivalence point is $50 \text{ cm}^3 + 50 \text{ cm}^3 = 100 \text{ cm}^3 = 0.100 \text{ dm}^3$.

The concentration of the conjugate base, acetate ion ($\text{CH}_3\text{COO}^-$), is:
$$ [\text{CH}_3\text{COO}^-] = \frac{0.005 \text{ moles}}{0.100 \text{ dm}^3} = 0.05 \text{ M} $$

The acetate ion hydrolyzes in water, making the solution basic:
$$ \text{CH}_3\text{COO}^-(\text{aq}) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{COOH}(\text{aq}) + \text{OH}^-(\text{aq}) $$

We need to use the base dissociation constant ($K_b$) for the acetate ion. We can find this from the given $K_a$ and the ion product constant for water ($K_w$):
$$ K_w = K_a \times K_b $$ $$ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} $$

Using the equilibrium expression for the hydrolysis:
$$ K_b = \frac{[\text{CH}_3\text{COOH}][\text{OH}^-]}{[\text{CH}_3\text{COO}^-]} $$
Let $x = [\text{OH}^-]$ at equilibrium. We can assume $[\text{CH}_3\text{COO}^-]$ is approximately $0.05 \text{ M}$.
$$ 5.56 \times 10^{-10} = \frac{(x)(x)}{0.05} $$ $$ x^2 = (5.56 \times 10^{-10})(0.05) = 2.78 \times 10^{-11} $$ $$ x = \sqrt{2.78 \times 10^{-11}} = 5.27 \times 10^{-6} \text{ M} $$
So, $[\text{OH}^-] = 5.27 \times 10^{-6} \text{ M}$.

Now, we can find pOH and then pH:
$$ \text{pOH} = - \log_{10}[\text{OH}^-] = - \log_{10}(5.27 \times 10^{-6}) \approx 5.28 $$ $$ \text{pH} = 14 - \text{pOH} = 14 - 5.28 = 8.72 $$
The pH of the solution at the equivalence point is approximately **8.72**.

Explanation of the Titration Curve's Shape

The titration curve for this reaction will have a characteristic shape due to the weak acid and strong base combination.

• Initial Region: The pH starts at a value greater than 1 (acidic, but less than a strong acid with the same concentration). The initial pH is calculated from the dissociation of the weak acid.
• Buffer Region: As $\text{NaOH}$ is added, it reacts with the acetic acid to form acetate ions. The solution now contains a significant amount of both the weak acid ($\text{CH}_3\text{COOH}$) and its conjugate base ($\text{CH}_3\text{COO}^-$), forming a **buffer solution**. This is the key feature of the curve, where the pH changes very little with the addition of the titrant. The pH in this region can be calculated using the Henderson-Hasselbalch equation. The midpoint of this region is where $pH = pK_a$.
• Equivalence Point: This is the point where all the acetic acid has been neutralized. As calculated above, the resulting solution contains only the conjugate base (acetate), which hydrolyzes water to produce a basic solution. This causes the equivalence point to occur at a **pH greater than 7**. The change in pH around the equivalence point is sharp, but less vertical than a strong acid-strong base titration.
• Post-Equivalence Point: Once the equivalence point is passed, the solution contains excess strong base ($\text{NaOH}$). The pH is now determined by the concentration of the strong base, and it will increase more gradually until it levels out at a high pH.

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