Pre-Laboratory Assignment #5
Measurement of pH
Name: ______________________________
1.
- 0.10 M HCl
HCl strips completely into H+ and Cl- ions in water. HCl-→ H+ +Clˉ
As HCl is a leading strength acid, it dissociates completely, and there will be the proper equilibrium between H+ ions concentration and the initial concentration of HCl. pH = -log[H⁺] = -(-1) = 1 = 1 which means pH is 1.
- 0.10 M NaOH
When this salt NaOH dissociate in water the product is Na+ and OH- ions. NaOH → Na⁺ + OH⁻.
Because of the strong character traits of NaOH, it completely decomposes to form OH⁻ ions which is the same as that of the initial concentration of NaOH. pH = pOH = -log [OH⁻] = -log (0,1) = -(-1) = 1 pH = 14 – pOH = 14 – 1 = 13 - 0.10 M HCOOH
Note that HCOOH is a weak acid and to some extent gets hydrated to form H⁺ and HCOO⁻ ions in water. HCOOH <–> H+ HCOO-.
1.8 x 10′-⁴ is given as the K for the formation of formic acid (HCOOH) in the chemical reaction. We will consider H^ {+} ions population to be denoted by the symbol x. Ka = H ⁺ HCOO⁻ HCOOH 1.8 x 10 ⁴ = x * x ÷ (0.10 – x).
Since the dissociation is very small, the x can be taken as 0.10 in against to its smallness. 1.8 × 10⁻^ {4} ≈ (0.1) x² ≈ 1.8 x 10⁻⁵ ⇒ x ≈ √(1.8×10⁻⁵) = 1.34 x 10⁻³
pH = -log[H⁺] = -log (1.34 x 10⁻³) = -(-2.87) ≈ 2.87 = 2.87 - 0.10 M NaHCO₂
NaHCO₃ disintegrates in water thereby leading to formation of Na⁺, HCO₃⁻, and OH⁻ ions. NaHCO₂ → Na⁺+HCO₃⁻+OH⁻.As the OH⁻ concentration is of little importance relative to the HCO₃⁻ concentration, in the academic paper we can lose it. H+ + CO₃⁻⁻ = H+ → NaOH + CO₂ Let’s assume x is the amount of HCO₃⁻ ions. Ka = [H⁺][HCO₃⁻] / [NaHCO₃] 1.8 x 10⁻⁴ = ((0.10 – x) * x)/x => x = 0.00317
- 0.10 M NH₄Cl NH₄Cl forms dissociated ions NH₄⁺ and Cl⁻ in water. NH₄CL →NH₄⁺+Cl⁻.
The NH₄⁺ ion exists as the result of dissociation of a weak acid H⁺ and a base NH₃ to produce more protons. NH₄⁺ ⇌ NH₃ + H⁺ is the Kb and is given as 1.8 x 10⁻⁵. Let x denoting the concentration of H⁺ ions. The dissociation of an amine takes place at the acidic surface of the catalyst particle according to the following equation: Kb = [NH₃][H⁺] / [NH₄⁺] 1.8 x 10⁻⁵ = x x x / (0.10 – x)
With the relatively small dissociation, this deems x to be up against with 0.10. So, x < 0.10. 1.8 x 10⁻⁵ = 1.8 x 10⁻⁶ x ≈ ? (1.8 x 10⁻⁶)⁰½ ≈ ? ≈ 4.24 x 10⁻⁴
pH = -log[H⁺] = -log (4.24 × 10⁻⁴) ≈ 3.37 ⇒ 3.37 - 0.10mol HCOOH / 0.10mol NaHCO3 buffer
Since we have equal concentrations of HCOOH and HCOO⁻ (from NaHCO₂), we can use the Henderson-Hasselbalch equation to calculate the pH: pH + log[A-]/[HA] = pKa= -log (Ka) = -10⁻⁴ = ~3.74
pH = 3.74 + log (0.10 + 0.10) = 3.74 + log(1.0) = 3.74.
2. I predict the pH of 0.10 M hydrochloric acid to be lower than that of 0.10 M formic acid solution because hydrochloric acid is a strong acid that, completely dissociates in the water to form a high concentration of hydrogen ions. Unlike formic acid, which mostly remains un-dissociated at neutral pH levels, hydrochloric acid completely disintegrates to give rise to high concentrations of hydrogen ions and therefore lower pH values at the same molarity.
3. The pH of 0.10 M ammonium chloride is predicted to turn out lower than the pH of 0.10 M sodium chloride because ammonium chloride is a salt composed of a weak base and a strong acid. When ammonium chloride is dissolved in water, hydrolysis prompts it to react with the water and give rise to ammonia (NH₃) and hydronium ions (H₃O⁺). It means that hydronium ion concentration is rising and pH is being lowered although, in the case of sodium chloride, hydrolysis undergoes at only a small level.
4. With the ready addition of a strong acid, a solution with the portion 0.10 M formic acid (HCO2H) and 0.10 M sodium formate (NaHCO2) prevents changes in pH due to its bullying nature. A buffer’s solution is a mixture of a weak acid and its conjugate base (alternatively a weak base and its conjugate acid) in relatively equal concentrations. Moreover, in case a powerful acid is introduced in the buffer mixture, the weak acid of the buffer solution to the acid neutralizes the hydrogen ions thus keeping the pH relatively unchanged. Hence, in this case, the formic acid which is the weak acid, and the sodium formate which is an ionic associate of sodium and formate work together to provide the solution the resistance against a change in pH when a strong acid is added.
