1. The pH of a 0.10 HCN solution is 5.2. At equilibrium, a. What is the [H30*] i
ID: 1031303 • Letter: 1
Question
1. The pH of a 0.10 HCN solution is 5.2. At equilibrium, a. What is the [H30*] in that solution. b. What is [CN ]? c. What is the [HCN]? d. What is the value of Ka for HCN? 2. For the solution above, what is the [OH] at equilibrium? 3. Degree of Ionization(a) In calculating for the disassociation above, there are two options for the numerator of the formula. What are those two options? a. concentration of or concentration of i. Are these initial concentrations or at equilibrium? b. What is the denominator? concentration of i. Is this an initial concentration or at equilibrium? The looks like a % calculation, but it's a ratio. Remember that the difference is that you don't multiple by 100' So, what is the value of (calculate the value) C. 4. Finish the following Acid/Base reactions KCN+H20 + Kt(aq) a. (Hint: Nitrogen-containing positive, polyatomic ions (similar to NH) are usually weak acids. Also, don't change the #H after the carbon)Explanation / Answer
a)
use:
pH = -log [H3O+]
5.2 = -log [H3O+]
[H3O+] = 6.31*10^-6 M
Answer: 6.31*10^-6 M
b)
HCN dissociates as:
HCN + H2O -----> H3O+ + CN-
0.1 0 0
0.1-x x x
SO,
[H3O+] = x = 6.31*10^-6 M
[CN-] = x = 6.31*10^-6 M
Answer: 6.31*10^-6 M
c)
[HCN] = 0.10 - x
= 0.10 - (6.31*10^-6)
= 0.10 M
Answer: 0.10 M
D)
Ka = [H+][CN-]/[HCN]
Ka = (6.31*10^-6)*(6.31*10^-6)/(0.10)
Ka = 3.98*10^-10
Answer: 3.98*10^-10
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