Question

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 K_a for this acid?

2. A weak monoprotic acid has pK_a = 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 (K_a = 1.8 x 10^-5) and 0.05 M in HCN (K_a = 6.2 x 10^-10)

4. Carbonic acid, H₂CO₃, is a diprotic acid with K_a1 = 4.2 x 10^-7 and K_a2 = 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

Answer 1

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