Question

Consider the following four titrations (i-iv): i. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M} \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) by \(0.2 \mathrm{M} \mathrm{HCl}\) ii. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M}\) HCl by \(0.2 \mathrm{M} \mathrm{NaOH}\) iii. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M} \mathrm{HOCl}\left(K_{\mathrm{a}}=3.5 \times 10^{-8}\right)\) by \(0.2 \mathrm{M} \mathrm{NaOH}\) iv. \(150 \mathrm{mL}\) of \(0.2 \mathrm{M} \mathrm{HF}\left(K_{\mathrm{a}}=7.2 \times 10^{-4}\right)\) by \(0.2 \mathrm{M} \mathrm{NaOH}\) a. Rank the four titrations in order of increasing \(\mathrm{pH}\) at the halfway point to equivalence (lowest to highest \(\mathrm{pH}\) ). b. Rank the four titrations in order of increasing \(\mathrm{pH}\) at the equivalence point. c. Which titration requires the largest volume of titrant (HCl or \(\mathrm{NaOH}\) ) to reach the equivalence point?

Short answer

a. The pH at the halfway points, from lowest to highest, are: ii - HCl: pH = 1, iv - HF: pH = 3.14, iii - HOCl: pH = 7.46, i - NH3: pH = 9.26. b. The pH at the equivalence points, from lowest to highest, are: ii - HCl + NaOH: pH = 7 (neutral), i - NH3 + HCl: pH slightly higher than 7, iv - HF + NaOH: pH higher than 7, iii - HOCl + NaOH: pH highest among these titrations. c. All four titrations require the same volume of titrant to reach the equivalence point, as their initial moles of analyte are the same (150 mL * 0.2 M).

Step-by-step solution

Find the pH at the halfway point for all titrations

We begin by calculating the pH at the halfway point for all four titrations. A general rule to remember is that at the halfway point of a titration involving weak acids or bases, the pH = pKa for weak acids or pH = 14 - pKb for weak bases. For strong acids or bases, we can find the pH using -log[H+] or pOH = -log[OH-]. i. For ammonia (a weak base), the pKb = -log(1.8*10^-5) = 4.74. So, the pH = 14 - 4.74 = 9.26 ii. For HCl (a strong acid) with [H+] = 0.1M (since only half of 0.2M HCl is consumed at halfway point), the pH = -log(0.1) = 1 iii. For HOCl (a weak acid), the pKa = -log(3.5*10^-8) = 7.46. So, the pH = 7.46 iv. For HF (a weak acid), the pKa = -log(7.2*10^-4) = 3.14. So, the pH = 3.14 Ranking these, we find the order from lowest to highest is ii, iv, iii, i.

Find the pH at the equivalence point for all titrations

The pH at the equivalence point depends on whether the original reactant being titrated was a weak acid or weak base and what its conjugate is (acidic, basic, or neutral). i. Ammonium ion (NH4+) is the conjugate acid of NH3 and it behaves as a weak acid. To calculate the pH, we would have to solve a quadratic equation using the equilibrium constant for its reaction with water. ii. NaCl, the product of this titration, is a salt of a strong base (NaOH) and a strong acid (HCl). Therefore, it is neutral and the pH at the equivalence point equals 7. iii. The product of this titration is the salt NaOCl, which is the sodium salt of hypochlorous acid. OCl- will hydrolyze in water and raise the pH above 7. iv. Sodium fluoride (NaF) is the product of this titration. Fluoride ion is the conjugate base of the weak acid HF and will act as a weak base in water, so the pH at the equivalence point will be greater than 7. Therefore, the order in terms of increasing pH at equivalence point is ii, i, iv, iii.

Determine which titration requires the largest volume of titrant

The volume of titrant required to reach the equivalence point is equal in all cases because all initial moles of analyte (acid or base being titrated) are the same (150mL * 0.2M). The equivalence point occurs when the initial moles of analyte equal the moles of titrant added. Hence, all four titrations require the same volume of titrant to reach the equivalence point.

Key concepts

Acid-Base Titration

Acid-base titration is a laboratory technique used to determine the concentration of an acid or a base in a solution. In this method, a solution of known concentration (the titrant) is slowly added to a measured volume of a solution with an unknown concentration (the analyte) until the reaction reaches the point where the added titrant stoichiometrically neutralizes the analyte; this is known as the end point. To accurately find this point, indicators that change color at or near the desired pH are commonly employed, or pH meters may be used for more precise measurements.

For students grappling with this concept, it is important to visualize titration as a process of neutralizing an acid with a base (or vice versa) until equivalence is achieved. At the start, the pH of the solution depends on the nature of the analyte. As the titrant is added, the pH changes in a predictable manner, usually shown on a titration curve, which plots pH against the volume of titrant added. Different types of acids and bases (strong vs. weak) will exhibit different shapes of titration curves.

Equivalence Point

The equivalence point in a titration is the exact point at which the amount of titrant added is stoichiometrically equal to the amount of analyte in solution. For instance, it is the moment when all the acid (or base) has reacted with the added base (or acid). The pH at the equivalence point can vary depending on the strength of the reactants.

For students, it is crucial to note that the equivalence point does not always mean the pH will be neutral (7.0). In an acid-base titration involving a strong acid and a strong base, the equivalence point does indeed have a pH of 7. However, when a weak acid is titrated with a strong base, the pH at the equivalence point will be greater than 7 because the salt and water produced give rise to a solution with basic properties. Conversely, when a weak base is titrated with a strong acid, the pH will be less than 7. Knowing the type of acids and bases that are reacting is key to predicting and understanding the pH at the equivalence point.

Halfway Point Titration

The halfway point in a titration is of considerable importance, especially when dealing with weak acids or bases. At this juncture of the titration process, exactly half of the analyte has been neutralized. This fact can be harnessed for its practical implications: for weak acids, the pH at this point is equal to the pKa, the negative log of the acid dissociation constant, and for weak bases, the pH is equal to 14 minus the pKb, the negative log of the base dissociation constant.

For students, understanding the significance of the halfway point can simplify the solving of titration problems. It can elucidate the properties of the analyte in a reaction, as at the halfway point, the concentrations of the acid and its conjugate base (or base and its conjugate acid) are equal, resulting in a buffer solution where the pH is most resistant to changes in the presence of added acids or bases. The simplicity of calculating the pH at this point makes the halfway juncture vital for titration calculations and in-depth studies of acid-base chemistry.