A 0.10 M solution of a weak monoprotic acid has a hydronium-ion concentration of 5.0 ´ 10-4 M. What is the equilibrium constant, Ka, for this acid?
a. 5.0 ´ 10-2
b. 5.0 ´ 10-3
c. 2.5 ´ 10-4
d. 2.5 ´ 10-5
e. 2.5 ´ 10-6
2. (Points: 1)
What is the carbonate-ion, CO3-2, concentration in a 0.037 M carbonic acid solution?
a. 1.2 ´ 10-4
b. 4.2 ´ 10-7
c. 7.6 ´ 10-8
d. 4.8 ´ 10-11
e. 5.2 ´ 10-19
3. (Points: 1)
Which solution would show the GREATEST change in pH on the addition of 10.0 mL of 1.0 M NaOH to 1.0 L of each of the following solutions?
a. 0.50 M CH3COOH + 0.50 M NaCH3COO.
b. 0.10 M CH3COOH + 0.10 M NaCH3COO.
c. 0.50 M CH3COOH.
d. 0.10 M CH3COOH.
e. 0.50 M NaCH3COO.
4. (Points: 1)
Which of the following, when added to water, will not change the pH?
a. NaHCO3
b. NH4NO3
c. K2SO3
d. KI
e. HCN
can u please show me working out, i am in emergency.
A 0.10 M solution of a weak monoprotic acid has a hydronium-ion concentration of 5.0 ´ 10^-4 M. What is the equilibrium constant, Ka, for this acid?
a. 5.0 ´ 10^-2
b. 5.0 ´ 10^-3
c. 2.5 ´ 10^-4
d. 2.5 ´ 10^-5
e. 2.5 ´ 10^-6
2. (Points: 1)
What is the carbonate-ion, CO3^-2, concentration in a 0.037 M carbonic acid solution?
a. 1.2 ´ 10^-4
b. 4.2 ´ 10^-7
c. 7.6 ´ 10^-8
d. 4.8 ´ 10^-11
e. 5.2 ´ 10^-19
too many questions in one post. Pick one or two that represent types and repost.
Question 3
Answer: B
Sure! I'll explain how to solve each of these questions step by step.
Question 1: The equilibrium constant, Ka, is related to the concentration of the hydronium ions and the concentration of the acid. The acid is weak, which means it only partially ionizes in water. Let's say the acid is represented as HA, and it partially dissociates into H+ and A-. The equilibrium constant expression would be Ka = [H+][A-]/[HA].
Given that the hydronium-ion concentration is 5.0 ´ 10-4 M and the acid concentration is 0.10 M, you can substitute these values into the equilibrium constant expression. Ka = (5.0 ´ 10-4)(5.0 ´ 10-4)/(0.10) = 2.5 ´ 10-5.
Therefore, the correct answer is d. 2.5 ´ 10-5.
Question 2: Carbonic acid, H2CO3, is a weak diprotic acid that partially dissociates into bicarbonate ion, HCO3-, and carbonate ion, CO3-2. The concentration of the carbonate ion can be found by using the equilibrium constant expression for the second dissociation of carbonic acid.
The equilibrium constant expression for the second dissociation of carbonic acid is K2 = [CO3-2][H+]/[HCO3-]. Since the concentration of carbonic acid is given as 0.037 M, you can assume that the concentration of HCO3- is negligible compared to CO3-2. Therefore, the expression simplifies to K2 = [CO3-2][H+].
The concentration of carbonic acid, [H2CO3], is equal to the concentration of H+ because both are in a 1:1 ratio. Therefore, [H+] = 0.037 M.
Now, substitute the values into the expression for K2: K2 = [CO3-2] * 0.037 M.
Since the concentration of CO3-2 is what we're trying to find, rearrange the equation to solve for [CO3-2]: [CO3-2] = K2 / 0.037 M.
Given that the typical value for K2 is approximately 4.8 ´ 10-11, substitute this value into the equation: [CO3-2] = (4.8 ´ 10-11) / 0.037 M ≈ 1.3 ´ 10-9.
Therefore, the correct answer is (rounded) e. 1.3 ´ 10-9.
Question 3: To determine which solution would show the greatest change in pH upon the addition of NaOH, we need to compare the initial concentrations of the common ion, which is CH3COO- in this case.
The solution that will show the greatest change in pH will have the highest initial concentration of CH3COO-. Comparing the options, we can see that solution a. 0.50 M CH3COOH + 0.50 M NaCH3COO has the highest concentration of the common ion.
Therefore, the correct answer is a. 0.50 M CH3COOH + 0.50 M NaCH3COO.
Question 4: When a substance is added to water and does not change the pH, it means the substance is a neutral salt formed from the reaction between a strong acid and a strong base.
Among the options, NaHCO3 is the only substance that fits this description. NaHCO3 is a salt formed from the reaction between a strong acid (HCl) and a strong base (NaOH).
Therefore, the correct answer is a. NaHCO3.
I hope this helps! Let me know if you have any further questions.