Pre-lab Questions: 1. What is an acid/base indicator? List the color ranges, pH,

Question

Pre-lab Questions:

1. What is an acid/base indicator? List the color ranges, pH, and any other characteristics of three common acid/base indicators: methyl red, methyl orange, and phenolphthalein.

2. The concentration of commercial vinegar is 5% w/v. Calculate the concentration in terms of molarity

3. Which equation can be used to find the pH of a buffer?

4. Calculate the pH of a buffer containing 0.20 M CH3COOH and 0.20 M CH3COONa. What is the pH after adding 10.0 mL of 0.10 M HCl to 65.0 mL of the buffer?

Explanation / Answer

1. An acid/base indicator is a weak acid or base itself, used to figure out the end point of an acid-base titration by their colour change i.e. it shows different colour itself at different pH medium.

2. pH color change characteristic feature

Methyl red 4.8-6.0 yellow (acid), red (base) pKa = 5.1,used to identify bacteria producing stable acids

Methyl orange 3.2-4.4 red (acid), yellow (base) pKa = 3.7, changes colour at the pH of a midstrength acid, is usually used in titrations for acids

Phenolphthalein 8.2-10 colorless (acid), pink (base) pKa = 9.3 Ideal for titration of strong acids with strong bases

2. concentration of commercial vinegar is 5% w/v

i.e. 100 mL solution contains 5 g of vinegar

Hence, 1000 mL contains 50 g i.e. 50/60 mol = 0.83 mol [vinegar i.e. acetic acid (CH3COOH) has a molecular weight of 60]

i.e. Concentration of that solution is 0.83 M

3. Henderson Hasselbach equation is used to find the pH of a buffer. Buffer can be of 2 types:

weak acid mixed with its conjugate base

equation: pH = pKa + log [base]/[acid]  [Ka = acid dissociation constant]

or weak base mixed with its conjugate acid

equation: pH = pKb + log [acid]/[base] [Kb = base dissociation constant]

4. buffer contains 0.20 M CH3COOH and 0.20 M CH3COONa

hence pH = pKa + log [CH3COO-]/[CH3COOH] = -log Ka + log 0.20/0.20 = -log (1.8 * 10^-5) + 0 = 4.74

initial [H+] = 1.8 * 10^-5 M

Hence, 65.0 mL of the buffer contains 1.8 * 10^-5 * 0.065 mol of H+ = 1.18 * 10^-6 mol of H+

Adding 10.0 mL of 0.10 M HCl (strong acid, dissociates completely into H+) gives 0.01 mL * 0.10 M = 0.001 mol of H+

Hence, total moles of H+ = (1.18 * 10^-6 + 0.001) = 1.00118 * 10^-3

So, final pH = -log [H+] = -log [1.00118 * 10^-3 ] = 3.00

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