Assignment Calculate the pH, pOH, [H+], and [OH] for each of the following
solutions. Show all work. (10 Points)
Solution Composition of Starting Materials
1 0.10 M acetic acid
2. 5 mL 0.10 M acetic acid + 5 mL wa
3 1 mL 0.10 M acetic acid + 99 mL water
4 5 mL 0.10 M acetic acid + 5 mL 0.10 M hydrochloric acid
5 0.10 M phosphoric acid
6 0.10 M ammonia
7 0.10 M ammonium nitrate
8 50 mL 0.10 M ammonia + 50 mL 0.10 M ammonium nitrate
9 10 mL solution 8 + 6 mL water
10 10 mL solution 8 + 5 mL water + 1 mL 0.10 M hydrochloric acid
11 10 mL solution 8 + 6 mL 0.10 M hydrochloric acid
12 10 mL solution 8 + 5 mL water + 1 mL 0.10 M sodium hydroxide
13 10 mL 0.10 M acetic acid + 5 mL 0.10 M sodium hydroxide
14 10 mL 0.10 M ammonium nitrate + 5 mL 0.10 M sodium hydroxide
You may want to rethink this post. I doubt you will get any takers for a 2 hour work assignment. If you wish to repost with an explanation of exactly what you don't understand about ONE of the problems, please do so and someone will help you understand that concept.
To calculate the pH, pOH, [H+], and [OH-] for each solution, we need to use the formulas and equations related to acid-base chemistry.
The pH is a measure of the acidity or basicity of a solution and is defined as the negative logarithm of the hydrogen ion concentration [H+]. The pOH is the negative logarithm of the hydroxide ion concentration [OH-]. The pH and pOH are related by the equation pH + pOH = 14.
To calculate the [H+] and [OH-] concentrations, we need to use the ionization constant of water (Kw), which is equal to 1.0 x 10^-14 at 25°C. In an aqueous solution, the concentrations of [H+] and [OH-] ions multiplied together always equal Kw: [H+][OH-] = Kw.
For each solution, we will need to consider the dissociation of the acids and bases present.
Let's go through each solution and calculate the pH, pOH, [H+], and [OH-].
Solution 1: 0.10 M acetic acid
Acetic acid (CH3COOH) is a weak acid that partially dissociates in water.
CH3COOH (aq) ↔ CH3COO- (aq) + H+ (aq)
Since acetic acid is a weak acid, we can assume that the concentration of [H+] is significantly less than the initial concentration of 0.10 M acetic acid. Therefore, we can assume that the [H+] concentration remains at 0.10 M, while the [OH-] concentration is negligible.
pH = -log10[H+] = -log10(0.10) = 1
pOH = 14 - pH = 13
[OH-] = 10^(-pOH) = 10^(-13)
Solution 2: 5 mL 0.10 M acetic acid + 5 mL water
Since the mixture is half acetic acid and half water, the concentrations will be diluted by a factor of 2 compared to Solution 1.
[H+] = 0.10 M / 2 = 0.05 M
[OH-] = Kw / [H+] = (1.0 x 10^-14) / (0.05) = 2.0 x 10^-13
pH = -log10(0.05) = 1.30
pOH = 14 - pH = 12.70
Solution 3: 1 mL 0.10 M acetic acid + 99 mL water
Similar to Solution 2, the acetic acid is further diluted and the concentrations will be even lower.
[H+] = 0.10 M / 100 = 0.001 M
[OH-] = Kw / [H+] = (1.0 x 10^-14) / (0.001) = 1.0 x 10^-11
pH = -log10(0.001) = 3.00
pOH = 14 - pH = 11.00
Continue the calculations for the remaining solutions using the same principles and equations mentioned above.