1. A weak monoprotic acid has molar mass 180 g/mol. When 1.00 g of this acid is dissolved in enough water to obtain a 300 mL solution, the pH of the resulting solution is found to be 2.62. What is the value of Ka for this acid?
2. A weak monoprotic acid has pKa = 3.08. Calculate the percent ionization of a 0.35 M solution of this acid.
3. Calculate the pH of a solution that is 0.050 M in CH3COOH (Ka = 1.8 x 10-5) and 0.05 M in HCN (Ka = 6.2 x 10-10)
4. Carbonic acid, H2CO3, is a diprotic acid with Ka1 = 4.2 x 10-7 and Ka2 = 4.7 x 10-11. Calculate the concentration of carbonate ions in a 0.80 M solution of carbonic acid.
please answer all 4 questions
1)
mass(HA)= 1.00 g
use:
number of mol of HA,
n = mass of HA/molar mass of HA
=(1 g)/(1.8*10^2 g/mol)
= 5.556*10^-3 mol
volume , V = 3*10^2 mL
= 0.3 L
use:
Molarity,
M = number of mol / volume in L
= 5.556*10^-3/0.3
= 1.852*10^-2 M
use:
pH = -log [H+]
2.62 = -log [H+]
[H+] = 2.399*10^-3 M
HA dissociates as:
HA
-----> H+ + A-
1.852*10^-2
0 0
1.852*10^-2-x
x x
Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Ka = 2.399*10^-3*2.399*10^-3/(0.01852-2.399*10^-3)
Ka = 3.569*10^-4
Answer: 3.57*10^-4
Only 1 question at a time please
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