Question

Consider the titration of \(80.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\), by \(0.400\) \(M\) HCl. Calculate the pH of the resulting solution after the following volumes of \(\mathrm{HCl}\) have been added. a. \(0.0 \mathrm{~mL}\) d. \(40.0 \mathrm{~mL}\) b. \(20.0 \mathrm{~mL}\) e. \(80.0 \mathrm{~mL}\) c. \(30.0 \mathrm{~mL}\)

Short answer

a. pH = 13.70 b. pH = 12.10 c. pH = 11.47 d. pH = 7 e. pH = 1

Step-by-step solution

a. 0.0 mL HCl

For 0.0 mL of HCl, the concentration of hydroxide ions remains unchanged because no acid has been added yet. We know the initial volume and concentration, so we can calculate moles of \(\mathrm{OH}^{-}\) ions present in the solution. Moles of \(\mathrm{OH}^{-}\) = volume of \(\mathrm{Ba(OH)_2}\) × molarity of \(\mathrm{Ba(OH)_2}\) Moles of \(\mathrm{OH}^{-}\) = \(80.0 \times 10^{-3}\) L × \(0.100\) mol/L = \(0.008\) mol Since barium hydroxide has two hydroxide ions, moles of \(\mathrm{OH}^{-}\) are doubled. Moles of \(\mathrm{OH}^{-}\) = \(0.008\) mol × 2 = \(0.016\) mol The concentration of \(\mathrm{OH}^{-}\) in the solution can be calculated as: \[\mathrm{[OH^{-}]} = \frac{0.016}{80 \times 10^{-3}}\] = \(0.20\) M Now, we can find the pOH: pOH = -log[\(\mathrm{OH}^{-}\)] = -log(0.20) Next, we will find the pH using the relation: pH = 14 - pOH

b. 20.0 mL HCl

First, we need to find the moles of \(\mathrm{H^+}\) ions introduced by the addition of HCl. Moles of \(\mathrm{H^+}\) = volume of HCl × molarity of HCl Moles of \(\mathrm{H^+}\) = \(20.0 \times 10^{-3}\) L × \(0.400\) mol/L = \(0.008\) mol Now, we compare the moles of \(\mathrm{OH}^{-}\) and \(\mathrm{H^+}\). Moles of \(\mathrm{OH}^{-}\) remaining after neutralization = (\(0.016\) mol - \(0.008\) mol) = \(0.008\) mol Next, calculate the new concentration of \(\mathrm{OH}^{-}\): \[\mathrm{[OH^{-}]} = \frac{0.008}{(80.0 + 20.0) \times 10^{-3}}\] = \(0.08\) M Now find pOH = -log[\(\mathrm{OH}^{-}\)] and then pH = 14 - pOH.

c. 30.0 mL HCl

Moles of \(\mathrm{H^+}\) = \(30.0 \times 10^{-3}\) L × \(0.400\) mol/L = \(0.012\) mol We will again compare the moles of \(\mathrm{OH}^{-}\) and \(\mathrm{H^+}\). Moles of \(\mathrm{OH}^{-}\) remaining after neutralization = (\(0.016\) mol - \(0.012\) mol) = \(0.004\) mol Next, calculate the new concentration of \(\mathrm{OH}^{-}\): \[\mathrm{[OH^{-}]} = \frac{0.004}{(80.0 + 30.0) \times 10^{-3}}\] = \(0.034\) M Now find pOH = -log[\(\mathrm{OH}^{-}\)] and then pH = 14 - pOH.

d. 40.0 mL HCl

Moles of \(\mathrm{H^+}\) = \(40.0 \times 10^{-3}\) L × \(0.400\) mol/L = \(0.016\) mol At this point, the moles of \(\mathrm{OH}^{-}\) and \(\mathrm{H^+}\) are equal, meaning both are completely neutralized and the solution is at the equivalence point. Therefore, the pH will be 7.

e. 80.0 mL HCl

Moles of \(\mathrm{H^+}\) = \(80.0 \times 10^{-3}\) L × \(0.400\) mol/L = \(0.032\) mol Now, we have more moles of \(\mathrm{H^+}\) than \(\mathrm{OH}^{-}\). Find the remaining \(\mathrm{H^+}\) ions: Moles of \(\mathrm{H^+}\) remaining after neutralization = (\(0.032\) mol - \(0.016\) mol) = \(0.016\) mol Next, calculate the new concentration of \(\mathrm{H^+}\): \[\mathrm{[H^+]} = \frac{0.016}{(80.0 + 80.0) \times 10^{-3}}\] = \(0.10\) M Now, we'll find pH: pH = -log[\(\mathrm{H^+}\)] = -log(0.10)

Key concepts

Acid-Base Titration

Acid-base titration is a laboratory procedure used to quantify the concentration of an unknown acid or base solution by reacting it with a known volume and concentration of the opposite solution. This reaction is called neutralization, where an acid and a base combine to form water and a salt.

During a titration, a specific pH change occurs at the end of the neutralization, known as the equivalence point, which is often indicated by a color change of the indicator substance added to the solution. The pH at various stages of the titration can be calculated by knowing the amount of acid and base, their respective concentrations, and the volume of the titrant added. Proper calculation and analysis of pH changes provide essential information about the strength and concentration of the solutions involved.

Neutralization Reaction

A neutralization reaction occurs when an acid and a base react to form water and a salt. This type of reaction is central to the process of titration and can be represented by a simple equation:

Acid + Base → Salt + Water

For instance, when hydrochloric acid (HCl) is added to a barium hydroxide (Ba(OH)_2) solution, barium chloride (BaCl_2) and water (H_2O) are produced. The moles of hydrogen ions (H^+) from the acid and hydroxide ions (OH^-) from the base are critical in determining the completion of the reaction, significant for understanding the point of neutralization within a titration.

Hydroxide Ion Concentration

Hydroxide ion concentration ([OH^-]) is a measure of the amount of hydroxide ions in a solution. It's a critical factor in determining the pH of a solution, especially in the context of acid-base titrations.

The concentration of hydroxide ions can be found by using the formula:

[OH^-] = Moles of OH^- / Volume of Solution

Where the 'Moles of OH^-' represents the amount of hydroxide available in the solution, and 'Volume of Solution' is the total volume of the basic solution and any added titrant. pH can be calculated using the pOH, which is the negative logarithm of the hydroxide ion concentration, and the relation pH + pOH = 14. As the titration progresses, monitoring the changes in hydroxide ion concentration helps in tracking the approach towards the equivalence point.

Equivalence Point

The equivalence point in a titration is where the moles of acid equal the moles of base. At this juncture, the neutralization reaction is theoretically complete, and no excess of either the acid or the base is present in the solution.

The pH at the equivalence point can vary depending on the strength of the acids and bases involved. For a strong acid-strong base titration, the equivalence point pH is typically 7, as it is the neutral point on the pH scale. However, for combinations where either the acid or base (or both) is weak, the equivalence point pH can be less than 7 (acidic) or greater than 7 (basic). Determining the equivalence point is crucial for calculating the unknown concentration in titrations and it helps chemists and students to conclude about the neutralizing power of the solutions involved.