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Chapter 10: Problem 6

A weak acid \(\mathrm{HA}\left(\mathrm{p} K_{\mathrm{a}}=5.00\right)\) was titrated with \(1.00 \mathrm{M}\) \(\mathrm{KOH}\). The acid solution had a volume of \(100.0 \mathrm{~mL}\) and a molarity of \(0.100 \mathrm{M}\). Find the \(\mathrm{pH}\) at the following volumes of base added and make a graph of pH versus \(V_{\mathrm{b}}: V_{\mathrm{b}}=0,1,5,9,9.9,10,10.1\), and \(12 \mathrm{~mL}\).

Short Answer

Expert verified
Calculate moles and use pKa to find pH at each step. Plot these values to visualize the titration.

Step by step solution

01

Calculate Initial Moles of Acid

First, calculate the initial moles of the weak acid HA. Given the volume as 100 mL and molarity as 0.100 M, the initial moles of HA are calculated using \( n = C \times V \). Here, \( n = 0.100 \times 0.100 = 0.010 \) moles.
02

Calculate moles of Base at Different Volumes

For each volume of base added, calculate the moles of KOH using \( n = C \times V \). The concentration is 1.00 M. E.g., for 1 mL of base, \( n = 1.00 \times 0.001 = 0.001 \) moles.
03

Compute pH Before Base Addition (\(V_b = 0\) mL)

At \( V_b = 0 \) mL, the HA is in the form of a weak acid with no added base. Use the pKa to find pH using the formula, \( \text{[H}^+\text{]} = \sqrt{K_a \cdot C_0} \), where \( K_a = 10^{-5} \) and \( C_0 = 0.10 \). Hence, \( [H^+] = \sqrt{10^{-5} \times 0.10} = 10^{-3} \). So, \( \text{pH} = 3.0 \).
04

Calculate pH after Adding 1 mL of Base

Here, the base added is in lesser moles compared to the acid, forming a buffer. Use the Henderson-Hasselbalch equation, \( \text{pH} = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) \), to find the pH for \( V_b = 1 \) mL. Calculate the remaining \([HA]\) and existence of \([A^-]\) then find pH.
05

Determine pH at \(V_b = 5\) mL

Using the buffer equation again, calculate moles of \([A^-]\) formed and \([HA]\) remaining. Apply the formula. \[ \text{pH} = 5.00 + \log \frac{0.005}{0.005} = 5.00 \].
06

Approach Equivalence Point for \(V_b = 9, 9.9, 10.1\) mL

At the equivalence point where \( V_b = 10 \) mL, all HA has converted to A-. After the equivalence point, any excess KOH determines the pH. For \( V_b = 10 \), it's the "equivalence point", so solve for \[ \text{pH} = 7.00 \] due to the formation of water and salts. For \( V_b = 9 \) and \( V_b = 9.9 \), use buffer equations for slightly acidic solutions. For \( V_b = 10.1 \), there is slight excess base making the solution basic.
07

Calculate pH Beyond Equivalence Point at \(V_b = 12\) mL

Post-equivalence, excess OH- dictates the pH. Calculate excess moles of OH- added, remaining after neutralizing HA, then find the concentration of \(OH^-\) and thus pOH, converting to pH.
08

Plotting pH versus Volume of Base

Finally, plot these calculated pH values against the respective volumes of base added on the x-axis to visualize the titration curve. Observe the buffer region and sharp rise at equivalence.

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Key Concepts

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

Weak Acid
A weak acid is one that does not fully dissociate in water, meaning it releases H\(^+\) ions and anions slowly and incompletely. This partial dissociation leads to an equilibrium between the undissociated acid (HA) and its ions (H\(^+\) and A\(^-\)). For example, the weak acid HA, with a given pK\(_a\) of 5.00, indicates its K\(_a\) value is 10\(^{-5}\). This measure gives us how easily the acid dissociates and thus affects its behavior during titration. Understanding weak acids is crucial during titration:
  • They form buffer solutions when mixed with their salts.
  • They have a distinct equivalence point compared to strong acids.
Recognizing these properties helps us predict and interpret the pH changes when a weak acid is titrated against a strong base like KOH.
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a powerful tool used to estimate the pH of a buffer solution. This equation is written as:\[\text{pH} = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) \]This equation is particularly useful when dealing with weak acids and their conjugate bases because it allows you to calculate the pH of a solution when the concentrations of the acid and base are known.For example, in a titration scenario like ours, if you've partially neutralized a weak acid, the remaining acid ([HA]) and the newly formed salt ([A\(^-\)]) create a buffer. This equation provides:
  • A mathematical relationship between the pH of the solution and the concentrations of the acid and its conjugate base.
  • A clear cut method to predict how the pH will change with incremental base addition, prior to reaching the equivalence point.
This makes it indispensable for understanding and plotting the pH changes during a titration.
Equivalence Point
The equivalence point in titration is a crucial concept. It refers to the stage where the amount of titrant added is just enough to completely neutralize the analyte solution. For weak acids, this point is distinct because it is typically not pH-neutral, due to the presence of the weak acid's conjugate base, which is a stronger base than water.During this experiment:
  • At exactly 10 mL of KOH added, the equivalence point is reached for the weak acid HA.
  • Here, all HA is converted to A\(^-\), and any added KOH will contribute to a rising pH beyond this neutral point.
  • This transition marks a sharp change on the titration curve, visible as a steep rise or drop in the pH graph.
Understanding this point is key to accurately determining the concentration and limits of the acid or base being titrated.
pH Calculation
Calculating the pH during a titration involves understanding the chemistry at each stage of titration. At different volumes of base added, the pH is influenced by:
  • The initial acidity of the weak acid solution.
  • The volume of strong base added, like KOH.
  • The formation of buffer solutions when HA begins converting into A\(^-\).
Initially, the pH is calculated using the concentration of the weak acid and its acid dissociation constant (K\(_a\)). As the titration progresses and base is added:- Use the Henderson-Hasselbalch equation to calculate pH in the buffer region (when both HA and A\(^-\) are present).- At the equivalence point, the solution largely contains A\(^-\), and the pH is determined by the basic nature of this ion. - Beyond the equivalence point, the pH is driven by the excess OH\(^-\) from the base, which results in a high pH.This sequential approach to pH calculation helps to create a clear titration curve, reflecting the characteristic changes in pH throughout the process.

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Most popular questions from this chapter

At what point in the titration of a weak base with a strong acid is the maximum buffer capacity reached? This is the point at which a given small addition of acid causes the least \(\mathrm{pH}\) change.

What is meant by the leveling effect?

Calculate the \(\mathrm{pH}\) at \(10.0-\mathrm{mL}\) intervals (from 0 to \(100 \mathrm{~mL}\) ) in the titration of \(40.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) piperazine with \(0.100 \mathrm{M} \mathrm{HCl}\). Make a graph of \(\mathrm{pH}\) versus \(V_{\mathrm{a}}\)

A certain acid-base indicator exists in three colored forms: $$ \begin{aligned} &\begin{array}{lllll} & \mathrm{H}_{2} \mathrm{In} & \stackrel{\mathrm{p} K_{1}=1.00}{\rightleftharpoons} & \mathrm{HIn}^{-} & \stackrel{\mathrm{p} K_{2}=7.95}{\rightleftharpoons} & \mathrm{In}^{2-} \\ \lambda_{\max }=520 \mathrm{~nm} & & \lambda_{\max }=435 \mathrm{~nm} & \lambda_{\max }=572 \mathrm{~nm} \\ \varepsilon_{520}=5.00 \times 10^{4} & \varepsilon_{4.35}=1.80 \times 10^{4} & \varepsilon_{572}=4.97 \times 10^{4} \\ & \text { Red } & \text { Yellow } & \text { Red } \end{array}\\\ &\varepsilon_{4.35}=1.67 \times 10^{4} \quad \varepsilon_{520}=2.13 \times 10^{3} \quad \varepsilon_{520}=2.50 \times 10^{4}\\\ &\varepsilon_{572}=2.03 \times 10^{4} \quad \varepsilon_{572}=2.00 \times 10^{2} \quad \varepsilon_{435}=1.15 \times 10^{4} \end{aligned} $$ The units of molar absorptivity, \(\varepsilon\), are \(\mathrm{M}^{-1} \mathrm{~cm}^{-1}\). A solution containing \(10.0 \mathrm{~mL}\) of \(5.00 \times 10^{-4} \mathrm{M}\) indicator was mixed with \(90.0 \mathrm{~mL}\) of \(0.1 \mathrm{M}\) phosphate buffer \((\mathrm{pH} 7.50)\). Calculate the absorbance of this solution at \(435 \mathrm{~nm}\) in a \(1.00-\mathrm{cm}\) cell.

Calculate the \(\mathrm{pH}\) when \(25.0 \mathrm{~mL}\) of \(0.0200 \mathrm{M}\) 2-aminophenol have been titrated with \(10.9 \mathrm{~mL}\) of \(0.0150 \mathrm{M} \mathrm{HClO}_{4}\).
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