Question

\(10 \mathrm{ml}\) of \(0.200 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) is added gradually to \(25 \mathrm{ml}\) of \(0.200 \mathrm{M} \mathrm{NaOH}\). Calculate the \(\mathrm{pH}\) of the initial \(0.200 \mathrm{M} \mathrm{NaOH}\) solution and after each successive addition of \(5 \mathrm{ml}\) of acid. \(\mathrm{K}_{\mathrm{W}}=10^{-14}\)

Short answer

The initial pH of the 0.200 M NaOH solution is 13.7. After the first addition of 5 ml of 0.200 M \(HC_2H_3O_2\), the pH is 13.4. After adding the remaining 5 ml of 0.200 M \(HC_2H_3O_2\), the final pH will be calculated similarly.

Step-by-step solution

1. Initial pH of NaOH solution

To find the initial pH of the NaOH solution, we must initially calculate the concentration of OH- ions. NaOH is a strong base, so it dissociates completely in water: \[ NaOH \rightarrow Na^{+} + OH^{-} \] The concentration of OH- ions will be the same as the molarity of NaOH which is 0.200 M. Then, we can calculate the pOH and pH using the given \(K_W\): \[ pOH = -\log{[OH^{-}]} = -\log{0.200} \] Now, we calculate the pH by using \(K_W\): \[ pH + pOH = pK_W = 14 \] \[ pH = 14 - pOH \]

2. Finding successive addition of the acid

First, let's find the amount of moles of \(HC_2H_3O_2\) and \(NaOH\). For initial HC2H3O2: \[ n_{HC_2H_3O_2_0} = V \times M = (10ml)(0.200 M) = 2 x 10^{-3} mol \] For initial NaOH: \[ n_{NaOH_0} = V \times M = (25 ml)(0.200 M) = 5 x 10^{-3} mol \] For each incremental addition of \(5ml\) of the acid solution, the moles of \(HC_2H_3O_2\) added will be: \[ n_{add} = V \times M = (5 ml)(0.200 M) = 1 x 10^{-3} mol \]

3. Calculating pH after each addition of acid

We will calculate the change in moles of \(HC_2H_3O_2\) and \(NaOH\) after every addition of \(5ml\) of acid solution, then find the new concentrations and finally, the pH of the mixture. For 1st addition: \[ n_{HC_2H_3O_2_1} = n_{HC_2H_3O_2_0} + n_{add} = 2 x 10^{-3} + 1 x 10^{-3} = 3 x 10^{-3} mol \] \[ n_{NaOH_1} = n_{NaOH_0} - n_{add} = 5 x 10^{-3} - 1 x 10^{-3} = 4 x 10^{-3} mol \] Determine new concentration by adding total volume: \[ [OH^{-}] = \frac{n_{NaOH_1}}{(\mathrm{Initial\, NaOH\, volume})+(\mathrm{added\, first\, HC_2H_3O_2\, volume})}= \frac{4 x 10^{-3} mol}{30 ml}=0.133 M \] Calculate new pH as seen in previous steps: \[ pOH = -\log{[OH^{-}]} = -\log{0.133} \] \[ pH = 14 - pOH \] Repeat these calculations for each additional \(5ml\) of \(HC_2H_3O_2\) added until full \(10 ml\) of acid is added, and find final pH.

Key concepts

Acid-Base Titration

Acid-base titration is an analytical chemistry technique used to determine the concentration of an acid or base in a solution by reacting it with a base or acid of a known concentration. The point at which the reaction is complete is known as the equivalence point, and it can be determined with the help of indicators or pH meters.

In our exercise, the titration involved adding a weak acid, acetic acid (HC2H3O2), to a strong base, sodium hydroxide (NaOH). The initial and subsequent pH changes indicate progress towards the equivalence point. By monitoring these changes, we can analyze the reaction's stoichiometry and compute the concentration of the acid or base present in the solution. To improve understanding, remember that the amount of acid and base at the start of the titration and the incremental additions play a crucial role in calculating the pH at each step.

pOH and pH Relationship

The pOH and pH relationship is a fundamental concept in acid-base chemistry and is crucial for pH calculation chemistry. These two values are inversely related and sum up to a constant value, which for pure water at 25°C, is 14, as shown in the equation (pH + pOH = 14). This implies that as the pOH decreases, indicating an increase in hydroxide ion concentration ([OH⁻]), the pH increases, indicative of a more basic solution.

In the given exercise, we first calculated the pOH using the concentration of hydroxide ions provided and then derived the pH. Consistently applying this relationship allows students to transition between the hydroxide ion concentration and the hydrogen ion concentration insights.

Acid and Base Molarity

Understanding acid and base molarity is essential for solution stoichiometry, particularly in titrations. Molarity, expressed as moles of solute per liter of solution (M), determines the reaction ratio of acids and bases. For example, molarity helps in predicting how many moles of acid will react with how many moles of base.

In our textbook solution, molarities of both (NaOH) and acetic acid are used to calculate the number of moles present prior to and during the titration process. Moreover, knowing the initial volumes and molarities of reactants is pivotal for finding the pH at each addition of acid to the base. In each increment, molarity leads to the mole calculation, which in turn is essential for determining the subsequent pH.

Weak Acid-Strong Base Reaction

A weak acid-strong base reaction, like the one in our exercise involving acetic acid and (NaOH), is characterized by the incomplete ionization of the weak acid and complete ionization of the strong base. During titration, the strong base will neutralize the weak acid, leading to a mixture of the conjugate base of the weak acid and the remaining strong base. This reaction is dynamic and calculating pH requires an understanding of equilibrium principles.

After each incremental addition of the acid in our example, the moles of acid and the remaining moles of strong base must be recalculated to find the resulting concentration of (OH⁻) ions. Taking into account the partial dissociation of the weak acid is sometimes necessary for accurately determining the pH, especially when approaching the equivalence point. Students should practice these calculations to fully grasp the acid-base interaction dynamics.