chemistry quizz....

Question

1. Calculate the concentration of H⁺ions in a 0.62 M NaOH solution.
2. Calculate the pH of 2.8 * 10 ^-4 M Ba(OH)₂
3. The ka for benzoic acid is 6.5 * 10 ^-5. Calculate the pH of a 0.10 M benzoic acid solution.
4. How much NaOH ( in grams) is needed to prepare 546 mL of solution with a pH of 10.0?
5. Calculate the pH of a 0.42 M NH₄Cl solution.

Answer 1

Apparently these are homework problems and you should probably attempt them and then ask for assistance. Here are examples of how to do some of them, but you should try the others.

1). 0.62 M NaOH ==> 0.62 M Na+ + 0.62 M OH-

[H+][OH-] = 1x10-14

[H+] = 1x10-14 / 0.62 = answer

2). pH = -log [H+]

pOH = -log [OH-]

pH + pOH = 14

2.8x10-4 M Ba(OH)2 and Ba(OH)2 ==> Ba2+ + 2OH-

[OH-] = 2 x 2.8x10-4 M = 5.6x10-4 M OH-

pOH = -log 5.6x10-4 = 3.25

pH = 14 - 3.25 = 10.7

5). NH₄⁺ + H₂O ==> NH₃ + H₃O⁺

Ka NH₄⁺ = 5.6x10^-10 (looked it up since it wasn't given)

Ka = 5.6x10^-10 = [NH₃][H₃O⁺] / [NH₄Cl]

5.6x10^-10 = (x)(x) / 0.42 - x and assuming that x is small relative to 0.42 we can ignore it

5.6x10^-10 = x² / 0.42

x² = 2.35x10^-10

x = [H₃O⁺] = 1.53x10^-5 M (above assumption was valid as this is small compared to 0.42)

pH = -log 1.53x10^-5 = 4.81