Lab 3

pKa Acetic acid

4.76

What will the pH if you dilute 100 ml of 0.5 M NaOH to 1L with distilled water?

# of moles of NaOH = 0.5 M*0.1L = 0.05 mole Total volume after dilution = 1L (it includes 100 ml of 0.5 M NaOH) Now calculate new Molar concentration of the NaOH solution Concentration of diluted NaOH = 0.05/1 = 0.05 M pOH = -log[OH] = -log0.05pOH =-(-1.3) = 1.3 and the pH = 14 - 1.3 = 12.7

Good Buffer Qualities

-highly soluble in water and least soluble in organic solvents -pKa is least affected by the medium -have least chemical interaction with the medium -have minimum absorbance between wavelengths of 230 - 700 nm -maintain enzymatic and hydrolytic activity

How many mls of 12 M Acetic acid must be added to 500 ml of 0.5 M Sodium Acetate to prepare a buffer of pH 4.76? (pKa of Acetic Acid is 4.76)

0.0104 L or 10.4 mL 12 N Acetic Acid

A change of 1 pH unit represents a _______change in the hydrogen ion concentration.

10 fold

Determine the pH of a solution prepared by mixing a 15 ml of 0.085 M NaOH and 135 ml of distilled water.

11.93

pKa of NH4OH

= 9.26

What are the concentrations of HOAc (Acetic acid) and OAc‑ (Acetate) in a 0.2 M Acetate buffer, pH 5.0? The pKa for HOAc = 4.76

Acetate = 0.127 M and Acetic Acid = 0.073 M

What volume of Acetic acid (17.6 N) and what amount (g) of Sodium Acetate would you need to prepare 100 ml of 0.2 M buffer, pH 4.66 (pKa of Acetic acid = 4.76 and MW of Sodium Acetate = 136)

Acetic Acid = 17.6 N Sodium Acetate = 136; pH = 4.66Concentration = 0.2 M; pKa of Acetic acid = 4.76; Volume of buffer = 0.1 L The total number of moles needed = 0.1 L x 0.2 M = 0.02 moles pH = pKa + log [A‑]/ [HA] 4.66 = 4.76 + log [A‑]/ [HA]or log [A‑]/[HA] = 4.66 ‑ 4.76 or log [A]/[HA] = ‑0.1 Now take antilog of ‑0.1 = [A‑]/[HA] = 0.79 ...................... equation 1 Let [A‑] = X, then [HA] = 0.02 ‑ X Plug in the above values in equation 1, X = 0.0158 ‑ 0.79 XX + 0.79X = 0.01581.79 X = 0.0158X = 0.0158/1.79 = 0.00883 moleThus [A‑] = 0.00883 moleAmount of Sodium Acetate (g) = 0.00883 mole x 136 = 1.2 g [HA] = Acetic Acid = 0.02 ‑ 0.00883 = 0.0112 moleAcetic acid = 0.0112/17.6 = 0.000636 L or 0.636 ml

Which of the compound is made up of strong acid and weak base? a) Sodium acetate b) Ammonium chloride c) Ammonium hydroxide d) None of the above

B

What is the name of pH indicator you will use today for colorimetric method?

Bromothymol Blue

Buffer capacity:

Buffer capacity is usually defined as the number of equivalents of either H+ or OH- ions that is required to change the pH of a given volume of buffer by one pH unit.

List three methods of pH estimation that you will perform in pH Lab?

Colorimetric (pH paper), Visual estimation, and pH meter

In Henderson‐Hasselbalch equation, [A‐] stands for:

Conjugate Base

What statement(s) is true when acids and bases neutralize each other? a) The amount of H+ and OH‐ is the same b) Water is produced c) Salt is produced d) All of the above

D

Calculate the ionic strength of 0.15 M solution of Fe2(SO4)3.

I = 2.25

Ionic strength can be calculated using the following formula:

I= 1/2 n∑I CiZi^2 Ci = molar concentration of ion species iZi = number of charges on that ion

Given that 0.1 M solutions of acetic acid and sodium acetate, describe the preparation of a 1 L of sodium acetate buffer, pH 5.4.

Sodium Acetate = 0.814 L and Acetic Acid = 0.186 L

Prepare 3 L of 0.2 M Acetate buffer, pH 5.0 starting with Sodium Acetate (MW= 136) and 1 N Acetic Acid (pKa = 4.76). Calculate the amount of Sodium Acetate (g) and amount of Acetic acid (ml)

Sodium Acetate = 51.8 g and Acetic acid = 0.22 L

pH Meter

The pH meter is potentiometer that measures the potential generated between a glass pH-sensing electrode and a reference electrode (calomel)

Buffer

is a solution containing a mixture of a weak acid [HA] and its conjugate base [A-] which is capable of resisting substantial changes in the pH following addition of acidic or basic molecules.

Assume you have a 150 ml of 0.1 M HCl solution. You took 100 ml from it and added 400 ml of distilled water into it. What will be the pH of this new HCl solution?

pH = 1.7

Calculate the pH and pOH of 0.2 M solution of NaOH. Click

pH = 13.3 pOH = 0.7

A 500 ml buffer solution containing 0.2 M formic acid and 0.25 sodium formate is mixed with 100 ml of 0.05 M HCl. What will be the pH of this new solution?

pH = 3.82 (using pKa of formic acid 3.76)

What is the pH of a solution when 100 ml of 0.1 N NaOH is added to 150 ml of 0.2 M HOAc?

pH = 4.46

The pH of acetic acid was adjusted with concentrated NaOH, at what pH would you find an equal amount of acetic acid and its conjugate base? The pKa for acetic acid is 4.76

pH = 4.76

What is the pH of the buffer containing: 0.5 M sodium acetate and 1.0 M acetic acid?

pH = 4.76 + log 0.5/1= 4.76 - 0.301 = 4.46

Calculate the pH of a solution prepared by mixing 500 ml of 0.25 M Sodium Acetate with 250 ml 0.1 M HCl.

pH = 5.36

A 0.2 L of 0.05 M solution of dibasic (Na2HPO4) phosphate buffer is mixed with 0.16 L of 0.05 M monobasic (NaH2PO4) phosphate buffer. Calculate pH of the buffer solution

pH = 7.3 when you use pKa =7.2

Calculate the pKa of lactic acid, given that when the concentration of free acid is 0.01 M and lactate is 0.087 M and the pH is 4.8.

pH = pKa + log [A-] /[HA] Rearranging the above equation and solving for pKa, we obtainpKa = pH - log [A-] /[HA] = 4.8 - log (0.087/0.01)= 4.8 - log (8.7)= 4.8 - 0.94 pKa = 3.86

Calculate the pH of a mixture of 0.1 M acetic acid and 0.2 M sodium acetate. The pKa of acetic acid is 4.76.

pH = pKa + log [A-] /[HA] = 4.76 + log 0.2/0.1 = 4.76 + 0.301 = 5.06_______

What is the pH of a 10‑8 M HCl solution?

pH = ‑log[H+][H+] = 10‑7 (H2O) + 10‑8 (HCl)= 1 x 10‑7 + 0.1 x 10‑7 = 1.1 x 10‑7pH = ‑log (1.1 x 10‑7)= log 1/1.1 x 10‑7 = log 0.9 x 107 = log 0.9 + log 107 = ‑0.046 + 7 = 6.95

Calculate the ratio of the acetate and acetic acid required in a buffer of pH 5.3.

pH= pKa + log [A-] /[HA] or log [A-] /[HA] = pH - pkalog [A-] /[HA] = 5.3 - 4.76 = 0.54[A-] /[HA] = antilog 0.54 = 3.47