Question

Consider the titration of \(150.0 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HI by \(0.250 \mathrm{M}\) NaOH. a. Calculate the \(\mathrm{pH}\) after \(20.0 \mathrm{mL}\) of \(\mathrm{NaOH}\) has been added. b. What volume of NaOH must be added so that the \(\mathrm{pH}=\) \(7.00 ?\)

Short answer

a. The pH after 20.0 mL of NaOH has been added is approximately 1.24. b. 60.0 mL of NaOH must be added to reach a pH of 7.00.

Step-by-step solution

a. Calculate the pH after 20.0 mL of NaOH has been added.

First, we need to calculate the moles of HI and NaOH in the titration. Moles of HI: \(C_{HI} \times V_{HI} = 0.100M \times 150.0mL = 0.0150 \,mol\) Moles of NaOH: \(C_{NaOH} \times V_{NaOH} = 0.250M \times 20.0mL = 0.0050 \,mol\) Now let's calculate the remaining moles of HI after the reaction with 20 mL of NaOH: Moles of HI remaining = 0.0150 mol - 0.0050 mol = 0.0100 mol Now we find the new concentration of HI in the solution: \[C_{HI_{remaining}} = \frac{0.0100 \,mol}{(150.0 + 20.0) mL} = 0.0571 \,M\] Next, we calculate the pH of the solution using the hydrogen ion concentration: \[\mathrm{pH} = -\log_{10}(C_{HI_{remaining}}) = -\log_{10}(0.0571) \approx 1.24\]

b. What volume of NaOH must be added so that the pH= 7.00?

At pH = 7.00, the solution is neutral, and moles of H+ and OH- will be equal. We can set up a stoichiometric equation to find the volume of NaOH required. Moles of HI = Moles of NaOH \(C_{HI} \times V_{HI} = C_{NaOH} \times V_{NaOH_{required}}\) Solve for the required volume of NaOH: \(V_{NaOH_{required}} = \frac{C_{HI} \times V_{HI}}{C_{NaOH}} = \frac{0.100 \,M \times 150.0 \,mL}{0.250 \,M} = 60.0\, mL\) So, 60.0 mL of NaOH must be added to reach a pH of 7.00.

Key concepts

Hydrogen Ion Concentration

Understanding hydrogen ion concentration is crucial to calculating pH during an acid-base titration. When we dissolve an acid like HI (hydroiodic acid) in water, it dissociates to produce hydrogen ions \((H^+)\).
This is important because the concentration of these ions directly affects the pH of the solution. In the given exercise, we start with a known concentration of HI and, as it's titrated with \(NaOH\) (sodium hydroxide), we see these ions being neutralized.

Calculating the remaining \((H^+)\) ions after some of the acid has reacted with the addition of \(NaOH\) gives insight into the acidity of the solution. The fewer \((H^+)\) ions are left, the less acidic (higher pH) the solution becomes.
This is why, after adding 20 mL of \(NaOH\), the hydrogen ion concentration drops and directly influences the calculated pH.

Stoichiometry

Stoichiometry is the bridge between quantities of reactants and products in a chemical reaction. In a titration, it helps us understand how much of one substance reacts with a given amount of another.
For this exercise, the reaction is between HI and \(NaOH\). Both compounds react in a 1:1 ratio:
\[\text{HI} + \text{NaOH} \rightarrow \text{NaI} + \text{H}_2\text{O}\]
This means for each mole of hydrogen iodide present, one mole of sodium hydroxide is needed to neutralize it.

Using stoichiometry, we can calculate the moles of each reactant to understand how much is required for complete neutralization. From these calculations, we found out how much \(NaOH\) was needed to bring the pH to 7.
Keeping track of moles helps predict the outcome of the titration process and ensures accuracy in reaching the desired pH.

pH Calculation

pH is a numerical scale used to specify the acidity or basicity of a solution. Calculating pH during a titration involves understanding how the concentration of hydrogen ions changes as more titrant is added.
In the given example, after adding \(20 \, \text{mL}\) of \(NaOH\) to a \(150 \, \text{mL}\) solution of \(0.1 \, \text{M}\) HI, we calculated the concentration of hydrogen ions still present.

• First step: Find the moles of the remaining \(H^+\) ions, as some have been neutralized.
• Second step: Divide these moles by the total volume of the solution.
• Third step: Use the pH formula, \(\text{pH} = -\log_{10}([H^+])\), to find the solution's pH.

This process allows us to measure acidity and allows predictions about how continued additions of base will affect the solution.

Neutralization Reaction

A neutralization reaction occurs when an acid and a base react to form water and a salt. For acid-base titrations, it is vital to understand these reactions. Neutralization determines the end point or equivalence point of a titration.
In the case of HI and \(NaOH\), this reaction can be expressed as:
\[\text{HI} + \text{NaOH} \rightarrow \text{NaI} + \text{H}_2\text{O}\]

• The reaction involves the hydrogen ion from the acid combining with the hydroxide ion from the base.
• This combination results in the formation of water, with excess ions forming a neutral salt (sodium iodide in this case).

The goal of the titration is to achieve complete neutralization, where all \(H^+\) ions from the acid are neutralized by \(OH^-\) ions from the base. This point is marked by achieving a neutral pH of 7, ensuring all the acid has been effectively neutralized by the base.