Question

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

Short answer

To achieve a pH of 8.00, a volume of \(3.71L\) of \(0.0100M\) NaOH must be added to the \(1.00L\) of \(0.0500M\) HOCl solution.

Step-by-step solution

Find the Ka and pKa of HOCl

We know that for \(HOCl\), the \(pKa\) equals \(7.53\). Calculated from the equation: \(Ka = 10^{-pKa}\), So the \(Ka\) for \(HOCl\) is \(10^{-7.53}\).

Use the Henderson-Hasselbalch equation to calculate the ratio of [A-] / [HA]

We can rearrange the equation, knowing the pH and \(pKa\) values to solve for \([A^-]/[HA]\): \( [{A^-}]/[{HA}] = 10^{(pH - pKa)} = 10^{(8.00 - 7.53)}\)

Compute the ratio \([A^-]/[HA]\)

Solve the equation for the \([A^-]/[HA]\) ratio: \([A^-]/[HA] = 10^{(8.00 - 7.53)} = 2.985\)

Calculate the molarity of A- and HA

We can say that the molarity \([A^-]\) and \([HA]\) add together to make the original molarity of \(HOCl\). Hence: \([HA] + [A^-] = 0.0500M\) Then, using the ratio \([A^-] / [HA] = 2.985\): \([A^-] = 2.986[HA]\) Substituting \( [A^-] = 2.986[HA] \) in the equation \([HA] + [A^-] = 0.0500M\), we can calculate the new molarity of \( [A^-] \) and \( [HA] \).

Determine the volume of NaOH needed

Knowing that \(NaOH\) reacts with \(HOCl\) to produce \(OCl^-\), with the molar ratio between \(NaOH\) and \(OCl^-\) being 1 : 1, the molarity change of \(OCl^-\) is equal to the molarity of \(NaOH\). So, \([NaOH] \times Volume_{NaOH} =\Delta [OCl^-] = [OCl^-]_{after} - [OCl^-]_{before} = 0.0371M - 0M = 0.0371M\) Solving for the volume of \(NaOH\), we have: \(Volume_{NaOH} = [NaOH]^{-1} \Delta [OCl^-] = 0.0100^{-1}M^{-1} \times 0.0371M = 3.71L\)

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a formula that relates the pH of a buffer solution to the concentration of its acid and base components. In simpler terms, it provides a way to calculate the pH when you know the ratio of the concentration of the acid form \( HA \) and its conjugate base \( A^- \) in the solution.

Here's the Henderson-Hasselbalch equation: \[ pH = pKa + \log\frac{[A^-]}{[HA]} \]
This equation is derived from the chemical equilibrium constant for the acid dissociation reaction:\[Ka = \frac{[H^+][A^-]}{[HA]}\] By taking the negative logarithm of both sides and rearranging, this transforms into the Henderson-Hasselbalch equation.

To use it, first calculate the \( pKa \) from the known \( Ka \) value of the weak acid. Then, by inserting the desired pH and the \( pKa \) into the equation, you can solve for the ratio of \( [A^-]/[HA] \). In our problem, this step is critical in figuring out how much \( NaOH \) needs to be added to achieve a specific pH.

Acid-Base Titration

Acid-base titration is a laboratory technique used to determine the concentration of an acid or a base by neutralizing it with a base or an acid of known concentration, respectively.

Determining the endpoint of the titration, where equivalent amounts of acid and base have reacted, is crucial. This can be indicated by a color change in a pH indicator or by reaching the desired pH level measured by a pH meter.

In the example problem, the titration involves adding \( NaOH \) to \( HOCl \) until the desired pH is reached, making it a classic case of acid-base titration. This process changes the concentration ratio of \( HOCl \) and its conjugate base \( OCl^- \), ultimately affecting the pH of the solution. Understanding the mole-to-mole ratio between the acid and base in the reaction is necessary to determine the right volume of \( NaOH \) to be added.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction in a chemical process, resulting in no net change in the concentration of products and reactants over time. It is a dynamic state, meaning that the reactions continue to happen, but they balance each other out.

In the context of our problem, the equilibrium involves the weak acid \( HOCl \) dissociating into hydrogen ions \( H^+ \) and its conjugate base \( OCl^- \). The chemical equilibrium constant \( Ka \) quantifies the extent of the acid dissociation and is used in the Henderson-Hasselbalch equation to estimate pH at equilibrium.

Understanding equilibrium is paramount in predicting how the system will respond to external changes such as the addition of \( NaOH \). Le Chatelier's principle helps to predict that adding a base would shift the equilibrium to produce more \( OCl^- \) until the system reaches a new equilibrium state.

Molarity Concentration

Molarity concentration, often just called molarity, is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution, and its unit is mol/L or M.

Molarity is a key concept in chemistry because it allows chemists to quantify the concentration of substances and to determine stoichiometry in reactions. In titrations, knowing the molarity of the titrant (the reagent of known concentration) and the analyte (the substance being analyzed) is critical.

In our exercise, we need to determine the final concentration of the \( OCl^- \) ion to find out how much \( NaOH \) to add. The initial molarity of \( HOCl \) is given, and through calculations involving the molarity and the volume, we deduce the volume of \( NaOH \) required to reach the optimum pH. The relationship between the volume of the solution and the concentration of solutes is a direct application of the concept of molarity in the process of acid-base titration.