Question

What volume of \(0.0100 M\) NaOH must be added to \(1.00 \mathrm{~L}\) of \(0.0500 \mathrm{M} \mathrm{HOCl}\) to achieve a \(\mathrm{pH}\) of \(8.00 ?\)

Short answer

To achieve a pH of \(8.00\), add \(1.00 \times 10^{-4}~L\) of \(0.0100~M\) NaOH to \(1.00~L\) of \(0.0500~M~HOCl\).

Step-by-step solution

Calculate the pOH and the concentration of \(OH^-\) ions required

Since we are given the target pH, we can calculate the pOH and the concentration of hydroxide ions, \(OH^-\), needed to achieve this pH. $$pOH = 14 - pH = 14 - 8.00 = 6.00$$ Now, using the pOH, we can find the concentration of hydroxide ions present: $$[OH^-] = 10^{-pOH} = 10^{-6.00} = 1.00 \times 10^{-6}~M$$

Calculate the moles of \(OH^-\) ions required

Now we have the target \(OH^-\) concentration needed to achieve a pH of 8.00, and since the volume of the \(HOCl\) solution is given, we can compute the moles of \(OH^-\) ions required: $$ moles~of~OH^- = (1.00 \times 10^{-6}~M) \times (1.00~L) = 1.00 \times 10^{-6}~moles$$

Calculate the volume of \(0.0100 M\) NaOH required

Since we know the number of moles of \(OH^-\) ions required and the concentration of NaOH, we can use the following equation to calculate the volume of NaOH required: $$Volume~of~NaOH = \frac{moles~of~OH^-}{[OH^-]~in~NaOH}$$ Plugging in the values, we get: $$Volume~of~NaOH = \frac{1.00 \times 10^{-6}~moles}{0.0100 M} = 1.00 \times 10^{-4}~L$$ So, to achieve a pH of 8.00, \(1.00 \times 10^{-4}~L\) of \(0.0100~M\) NaOH should be added to \(1.00~L\) of \(0.0500~M~HOCl\).

Key concepts

pH and pOH relationship

Understanding the connection between pH and pOH is essential when working with solutions, especially when calculating buffer solutions.
The pH scale measures how acidic or basic a solution is. A pH of 7 is neutral, below 7 is acidic, and above 7 is basic. The pOH scale is complementary to the pH and represents the concentration of hydroxide ions in a solution.

• The sum of pH and pOH for any aqueous solution at 25°C is always 14. This relationship helps in converting pH to pOH and vice versa.
• Since pH = 8 was given, the calculation of pOH is straightforward:

\[ pOH = 14 - pH = 14 - 8.00 = 6.00 \]
This simple arithmetic demonstrates that if you know one, you can easily find the other. This is crucial in determining the concentration of hydroxide ions in the solution, guiding us further in the buffer solution calculations.

hydroxide ion concentration

Determining the hydroxide ion concentration is a key step in adjusting and preparing buffer solutions. Once the pOH is known, you can find the concentration of hydroxide ions (\([OH^-]\)), which is vital for calculating the necessary amount of base required for a desired pH.

• To convert from pOH to \([OH^-]\), use the formula:

\[ [OH^-] = 10^{-pOH} \]
In the exercise example, the pOH was calculated to be 6.00. Thus:

• \([OH^-] = 10^{-6.00} = 1.00 \times 10^{-6} \ M\)

This concentration represents the number of hydroxide ions per liter of solution required to achieve the target pH. Understanding this relationship is pivotal to ensuring accurate chemical measurements in buffer preparations.

NaOH solution preparation

Preparing a NaOH solution of a specific concentration involves careful calculation to ensure the correct amount of solution is added to achieve a desired pH. For the exercise at hand, where a specific volume of a NaOH solution is needed, the calculation focuses on both the concentration and the volume of the solution.

• First, determine the moles of \(OH^-\) required using the concentration of \([OH^-]\) and the volume of the original solution:

\[ moles\ of\ OH^- = [OH^-] \times volume \] In our scenario:

• \( moles\ of\ OH^- = 1.00 \times 10^{-6} \ M \times 1.00\ L = 1.00 \times 10^{-6}\ moles \)

To determine the volume of NaOH solution necessary, use:

• \( Volume\ of\ NaOH = \frac{moles\ of\ OH^-}{[OH^-]\ in\ NaOH} \)

Inserting the values calculated:

• \( Volume\ of\ NaOH = \frac{1.00 \times 10^{-6}\ moles}{0.0100\ M} = 1.00 \times 10^{-4}\ L \)

Such calculations are standard practice when preparing solutions where precise chemical reactions are critical. Accurately measuring and mixing chemicals can ensure experimental and practical success in various scientific fields.