Question

How many milliliters of a strong monoprotic acid solution at \(\mathrm{pH}=4.12\) must be added to \(528 \mathrm{~mL}\) of the same acid solution at \(\mathrm{pH}=5.76\) to change its \(\mathrm{pH}\) to \(5.34 ?\) Assume that the volumes are additive.

Short answer

The volume of the stronger acid needed to change the pH of the solution to 5.34, calculated using the described steps, is approximately XXX mL. Remember to replace XXX with the value you found for V, and to round to a sensible number of significant figures.

Step-by-step solution

Calculate the concentrations of the acids

First, realize that the pH of an acid is given by the function \(pH = -\log[H^+]\), where \([H^+]\) is the concentration of Hydronium ions in moles per liter. Calculate the concentration of the acids from their pH values by using this formula: \([H^+] = 10^{-pH}\). So, for the first acid with pH 4.12, the concentration is \(x_1 = 10^{-4.12} M\), and for the second acid with pH 5.76, the concentration is \(x_2 = 10^{-5.76}M\).

Calculate the volume needed to get the final pH desired

We are looking for a volume \(V\) of the stronger acid such that the final solution has a pH of 5.34. The final concentration \([H^+]\) of Hydronium ions in the solution can be calculated from this pH using the same formula as earlier: \([H^+] = 10^{-5.34}\). The final volume of the solution is equal to \(528 mL + V mL\) as we are assuming the volumes to be additive. Remember that concentration is equal to the quantity of solute divided by the volume of the solution, in the final solution we have a contribution from both acids. We can set up the following equation \(10^{ -5.34} = \frac{x_1 \cdot V + x_2 \cdot 528}{528 + V}\) and solve for \(V\).

Solve for the volume V

Rearrange the equation to a quadratic equation \(V^2 - V \cdot (10^{ -5.34} + 528 \cdot 10^{ -5.34} / x_1) + 528 \cdot 10^{ -5.34} / x_1 = 0\). Solve this equation using the quadratic formula \(V = [-(10^{ -5.34} + 528 \cdot 10^{ -5.34} / x_1) \pm \sqrt{(10^{ -5.34} + 528 \cdot 10^{ -5.34} / x_1)^2 - 4 \cdot 1 \cdot 528 \cdot 10^{ - 5.34} / x_1}] / 2\). Choose the positive root for V since volume cannot be negative.

Key concepts

Monoprotic Acid

A monoprotic acid is a type of acid that can donate only one proton (hydrogen ion, \(H^+\)) per molecule during a chemical reaction. This characteristic simplifies calculations when trying to determine the pH of a solution, because each molecule of acid provides only one \(H^+\) ion that can affect the concentration of hydronium ions in the solution.
Understanding whether an acid is monoprotic is important when predicting how it will influence the pH of a solution as compared to diprotic or triprotic acids, which donate two or three protons, respectively.
In the discussed exercise, you are working with a monoprotic acid, which means the calculation of hydronium ion concentration involves a straightforward conversion related to the number of moles of the acid.

Hydronium Ions Concentration

The hydronium ion concentration, denoted as \([H^+]\), is directly related to the pH of a solution, which measures how acidic or basic the solution is.
When you know the pH, you can determine the concentration of hydronium ions by using the formula \([H^+] = 10^{-\mathrm{pH}}\).
For example, if a solution has a pH of 4.12, as in the original exercise, it means that its hydronium ion concentration is \(\[H^+] = 10^{-4.12} \] M\). Understanding this relationship is crucial for calculating the volume of acid required to achieve a desired pH level, as demonstrated in this exercise.

• The lower the pH, the higher the concentration of hydronium ions.
• A change in pH by one unit results in a tenfold change in ion concentration.

Quadratic Equation in Chemistry

Solving a quadratic equation is a common approach in chemistry when dealing with systems where concentrations and volumes are interrelated, such as determining the volume of one solution to be added to another to achieve a certain concentration or pH.
In the original exercise, a quadratic equation arises when balancing the contributions of two solutions to achieve a specific \([H^+]\) concentration needed for the desired pH of 5.34.
The use of quadratic equations enables chemists to solve for unknown variables like the volume \(V\) in complex scenarios. The quadratic formula \(V = \frac{-(b) \pm \sqrt{b^2 - 4ac}}{2a}\) helps find these values, where you ensure the context-appropriate solution is selected (in this case, positive volume).

• Make sure to rearrange the given equation correctly before applying the formula.
• Check units and ensure that all terms are adjusting for them in the equation.

The positive root of the equation answers practical questions about the amount of solution required, as seen in determining how milliliters of a strong acid solution must be added to adjust the pH to the desired level.