Question

Consider the titration of a 35.0 \(\mathrm{mL}\) sample of 0.175 \(\mathrm{M} \mathrm{HBr}\) with 0.200 \(\mathrm{M} \mathrm{KOH} .\) Determine each quantity. a. the initial pH b. the volume of added base required to reach the equivalence point c. the pH at 10.0 \(\mathrm{mL}\) of added base d. the \(\mathrm{pH}\) at the equivalence point e. the pH after adding 5.0 \(\mathrm{mL}\) of base beyond the equivalence point

Short answer

a. The initial pH is 0.76, b. 30.6 mL of KOH is required to reach the equivalence point, c. The pH at 10.0 mL of added base is approximately 1.25, d. The pH at the equivalence point is 7, e. The pH after adding 5.0 mL of base beyond the equivalence point is approximately 12.30.

Step-by-step solution

- Calculate the initial pH

The initial pH of a solution of a strong acid like HBr can be found directly from its concentration by using the formula: \( \mathrm{pH} = -\log{[\mathrm{H}^+]} \). Since HBr is a strong acid that dissociates completely in water, \([\mathrm{H}^+] = [\mathrm{HBr}]\). Thus, \( \mathrm{pH} = -\log{(0.175)} \).

- Volume of base to reach the equivalence point

At the equivalence point, moles of HBr will equal moles of KOH. Use the formula \( M_\mathrm{acid} \times V_\mathrm{acid} = M_\mathrm{base} \times V_\mathrm{base} \) to find the volume of KOH required. Substituting the given values: \( 0.175 \mathrm{M} \times 35.0 \mathrm{mL} = 0.200 \mathrm{M} \times V_\mathrm{KOH} \). Solve for \( V_\mathrm{KOH} \).

- pH at 10.0 mL of added base

After adding 10.0 mL of the KOH base, calculate the moles of KOH added and the moles of HBr originally present. Subtract the lower value from the higher to find the remaining concentration and determine if the solution is acidic or basic. The pH if acidic is found by \( -\log{[\mathrm{H}^+]} \) or if basic by \( 14 + \log{[\mathrm{OH}^-]} \), since \( \mathrm{pOH} = -\log{[\mathrm{OH}^-]} \) and \( \mathrm{pH} + \mathrm{pOH} = 14 \).

- pH at the equivalence point

At the equivalence point, the number of moles of acid equals the number of moles of base, resulting in a neutral solution. However, since KOH is a strong base and HBr is a strong acid, the resulting solution is just water with a pH of 7 at 25°C because no excess ions are present to affect the pH.

- pH after adding 5.0 mL of base beyond the equivalence point

After reaching the equivalence point, any additional KOH will increase the pH of the solution. Calculate the number of moles of excess KOH, then divide by the total volume to find the molarity of \( \mathrm{OH}^- \). Use this to find the pOH using \( \mathrm{pOH} = -\log{[\mathrm{OH}^-]} \) and then the pH using \( \mathrm{pH} = 14 - \mathrm{pOH} \).

Key concepts

Equivalence Point Calculation

Understanding the equivalence point in an acid-base titration is crucial for interpreting the results. The equivalence point is the juncture in the titration process where the amount of acid neutralizes the amount of base, or vice versa.

For example, in the titration of hydrobromic acid (HBr) with potassium hydroxide (KOH), we calculate the equivalence point by using the formula:
\( M_{\mathrm{acid}} \times V_{\mathrm{acid}} = M_{\mathrm{base}} \times V_{\mathrm{base}} \). By inputting the acid's molarity and volume along with the base's molarity, we can solve for the volume of base needed to reach the equivalence point. Given the values in our exercise, we find the volume of KOH required.

Students often struggle with the concept of the acid and base reacting in a stoichiometric ratio. Remember, this is a 1:1 molar ratio since HBr and KOH each donate one hydrogen ion and one hydroxide ion, respectively, for the neutralization reaction.

pH Calculation

The pH scale is a measure of acidity or basicity of an aqueous solution. Calculation of pH is a vital step in acid-base titrations since it helps us monitor the progress and identify the end point.

To calculate the initial pH of a strong acid like HBr, we use the expression
\( \mathrm{pH} = -\log{[\mathrm{H}^+]} \) since the concentration of hydrogen ions, [H+], is equal to the acid's molarity due to complete dissociation in water. As we add the base, the pH will change - it increases until it reaches a value of 7 at the equivalence point if the titration is between a strong acid and a strong base at 25°C. Beyond the equivalence point, the solution becomes increasingly basic, and we use \( \mathrm{pH} = 14 - \mathrm{pOH} \) to find the pH, where \( \mathrm{pOH} = -\log{[\mathrm{OH}^-]} \).

• For an acidic solution: \( \mathrm{pH} = -\log{[\mathrm{H}^+]} \)
• For a basic solution: \( \mathrm{pH} = 14 - \mathrm{pOH} \)

Strong Acid and Base Reactions

Reactions between strong acids and strong bases are characteristically straightforward. Upon mixing, these substances ionize completely in aqueous solutions to produce hydrogen ions, H+, from the acid and hydroxide ions, OH-, from the base.

The reaction between a strong acid such as hydrobromic acid (HBr) and a strong base like potassium hydroxide (KOH) can be represented by a simple neutralization equation:
\( \mathrm{HBr} + \mathrm{KOH} \rightarrow \mathrm{KBr} + \mathrm{H}_2\mathrm{O} \).

The production of water and a salt, in this case, potassium bromide (KBr), is a hallmark of this type of reaction. There are no remaining excess hydrogen or hydroxide ions in the solution at the equivalence point, which makes the pH neutral, typically a pH of 7 at 25°C.
Also, understanding that the complete dissociation of these strong electrolytes is what drives the reaction to completion and produces predictable pH changes during the titration process is essential for students to grasp the concept of what happens during the titrations of strong acids with strong bases.