Need help with data sheets!!! 6 0499 Studying the pH of Strong Acid, Weak Acid,

Question

Need help with data sheets!!! 6 0499 Studying the pH of Strong Acid, Weak Acid, Salt, and Buffer Solutions Preyural by Nona Robinso North, State University of New York at Potsdam, an M. L. Gillette, Indiane Unioersity Kokome PURPOSE OF THE EXPERIMENT Compare calculated and measured pHs of a series of hydrochloric acid and acetic acid solutions. Based on the pH measurements, calculate the ald dissociation constant of acetic acid Measure the pH of various salt solutions. Using your pH meassure- ments, calculate the hydrolysis constant of ammonium chloride. Compare measured and calculated pHs of a buffer solution; the same buffer solution mixed with HCI solution and mixed with NaOH solution distilled water; and distilled water mixed with HCI solution and mixed with NaOH solution. BACKGROUND Water and Aqueous Solutions INFORMATION Many compounds dissolve in water, forming aqueous solutions. In some of these cases, water molecules simply surround, or solvate, molecules or ions in the solution. In other cases, the water molecules and the solute molecules or ions react. In pure water, a small percentage of the molecules take part in the chemical equilibrium shown in Equation 1. 2 HOU)HO (a)+OH (aq) As a result, small amounts of hydronium ion (H,O) and hydroxide ion (OH) are present in water. The equilibrium constant expression for this reaction, Kw., sometimes referred to as the ion product of water, is shown in Equation 2. We use square brackets to indicate the molar concentration (mol/L, M) of the enclosed species. Eq. 2) Learning

Explanation / Answer

To calculate the Ka of the acetic acid HC2H3O2 (literature Ka = 1.8 x 10-5)

remeber the dissociation reaction for a weak acid

HA ========= H + A

K = products / reactants

Ka = [H] *[ A] / [HA], ka is the equilibrium constant and the [H] indicates concentration of H and so on

so you say you have a ph of 2.7 PH = - log [H], [H] is the concentration of H

[H] = 10-2.7 = 0.00199, since the solution is left by it self concentration of A is equal to concentration of H

concentration of HA is the original concentration you already know for first trial 0.1

Ka = 0.00199 * 0.00199 / 0.1 = 0.000039601 or 3.96 x 10-5

For trial 2, PH = 3.2 and [acetic acid] = 0.01

H = 10-3.2 = 0.00063

Ka = 0.00063 * 0.00063 / 0.01 = 0.00003969 = 3.96 x 10-5

For trial 3, PH = 3.6 and ac. acid 0.001

H = 0.00025

Ka = 0.00025 * 0.00025 / 0.001 = 0.0000625 = 6.25 x 10-5

For trial 4, PH = 4.2 and ac. acid = 0.0001

H = 0.000063

Ka = 0.000063 * 0.000063 / 0.0001 = 0.00003969 = 3.96 x 10-5

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.