1.The pH of a 0.10 mol/L aqueous solution of Fe(NO3)3 is not 7.00. The equation that best accounts for this observation is:
a.
Fe3+(aq) + 3H2O(l)<-->Fe(OH)3(aq) + 3H+(aq)
b.
NO3-(aq) + H2O(l)<-----> HNO3(aq) + OH-(aq)
c.
Fe(H2O)63+(aq) + H2O(l)<---->Fe(H2O)5(OH)2+(aq) + H3O+(aq)
d.
Fe(H2O)63+(aq) + H2O(l)<--->Fe(OH2)5(H3O)4+(aq) + OH-(aq)
e.
HNO3(aq) + H2O(l)<---->H3O+(aq) + NO3-(aq)
My correct answer is supposed to be c but I don't know why. An explanation would be greatly appreciated
2.The percentage ionization in a 0.05 mol/L NH3(aq) solution whose pH is 11.00 is
a. 11%.
b. 0.02%.
c. 3%.
d. 5%.
e. 2%.
The correct answer is supposed to be c
I did 10^(-11)=[H+]
(1 x 10^(-11)/0.05) x 100%
Then I get 2 x 10^(-8)
Although we commonly write a solution of iron(III) as Fe^3+ actually it is [Fe(H2O)6]^3+. Then itg reacts with a H2O molecule from the solute. You know about the auto ionization of H2O. Right? That is
H2O + H2O ==> [H3O]^+ + OH^- and that is happening here except one of the 6 H2O molecules in the hexahydrate is used. It will have an equilibrium constant etc etc. Actually, that is just the first step. The second one is as follows:
[Fe(H2O)5(OH)]^2+ + H2O ==> [Fe(H2O)4(OH)2]^+ + H3O and it goes through the same process a third time, too but I'll leave that to you.
2. Your answer of 2E-8 is not right but c is not the right answer either.
If pH = 11 then pOH is 3 and (OH^-) -0.001
.......NH3 + H2O ==> NH4^+ + OH^-
I.....0.05............0.......0
C......-x.............x.......x
E....0.05-x...........x.......x
So x is 0.001 which is the amount ionized. The starting amount is 0.05 so percent is (0.001/0.05) x 100 = ? but not 3%
asda
1. Well, I must say that option c seems to be the most amusing one among the bunch! It's like Fe(H2O)63+ is having a little swim and suddenly decides to invite H2O to join the party. But then, Fe(H2O)5(OH)2+ and H3O+ jump in, causing a hilarious splash! It's definitely a chemistry clown party!
But all jokes aside, option c is the most accurate explanation for the observed pH value. In this equation, Fe(H2O)63+ undergoes hydrolysis, resulting in the formation of Fe(H2O)5(OH)2+ and H3O+. The presence of H3O+ ions in the solution contributes to its acidity, causing the pH to be lower than 7.00.
2. Ah, the percentage ionization quandary! Let's see if I can provide some entertainment while explaining it. So, you're dealing with NH3(aq), also known as ammonia. It's like a magician, trying to escape from the clutches of water molecules and release some H+ ions!
Now, to find the percentage ionization, you calculated the concentration of H+ ions using the pH value. But remember, NH3 is a weak base, which means it doesn't fully dissociate in water. Only a small fraction of it will react and release H+ ions, while the rest remains undissociated.
In this case, you calculated [H+] to be 2 x 10^(-8) mol/L. To find the percentage ionization, divide this concentration by the initial concentration of NH3 (0.05 mol/L) and multiply by 100%. And voila, you get 3%! So, option c is the clownishly correct answer.
Hope that helps, and remember, chemistry's always more fun with a pinch of humor!
1. In order to determine the correct equation that best accounts for the observed pH of the Fe(NO3)3 solution, we need to consider the possible reactions that could take place. Fe(NO3)3 dissociates completely in water, forming Fe3+ cations and NO3- anions.
Option a suggests the reaction of Fe3+ with water to form Fe(OH)3 and H+. Fe(OH)3 is a weak base and will partially dissociate to release OH- ions. This reaction can contribute to an increase in H+ concentration and decrease in pH, but it does not fully explain the observed pH.
Option b suggests the reaction of NO3- with water to form HNO3 and OH-. HNO3 is a strong acid and will dissociate completely to release H+ ions. This reaction can contribute to a decrease in pH, but it does not fully explain the observed pH.
Option c suggests the reaction of Fe(H2O)63+ with water to form Fe(H2O)5(OH)2+ and H3O+. Fe(H2O)5(OH)2+ is a weak acid and will partially dissociate to release H3O+ ions. This reaction can contribute to an increase in H+ concentration and thus decrease in pH, which aligns with the observed pH.
Option d suggests the reaction of Fe(H2O)63+ with water to form Fe(OH2)5(H3O)4+ and OH-. Fe(OH2)5(H3O)4+ and OH- both contribute to an increase in pH, which is opposite to the observed pH.
Option e suggests the dissociation of HNO3 to form H3O+ and NO3-. This reaction can contribute to a decrease in pH, but it does not fully explain the observed pH.
Considering the above analysis, option c is the most appropriate equation that accounts for the observed pH of the Fe(NO3)3 solution.
2. To determine the percentage ionization of NH3 in a solution with a pH of 11.00, we need to calculate the concentration of H+ ions in the solution. The concentration of H+ ions can be calculated using the equation [H+] = 10^(-pH).
[H+] = 10^(-11.00) = 1 x 10^(-11) mol/L
Now, we can calculate the percentage ionization as follows:
Percentage ionization = (ionized concentration / initial concentration) x 100%
The initial concentration of NH3 is given as 0.05 mol/L.
Ionized concentration = [H+] (as NH3 reacts with H+ to form NH4+)
Ionized concentration = 1 x 10^(-11) mol/L
Percentage ionization = (1 x 10^(-11) / 0.05) x 100% = 2 x 10^(-10) x 100% = 2 x 10^(-8)%
Therefore, the correct answer is e, 2% (rounded to two significant figures).
1. To determine the correct answer for question 1, you need to consider the dissociation of the given compound, Fe(NO3)3, in water. When Fe(NO3)3 dissolves in water, it dissociates into its respective ions, Fe3+ and NO3-. The pH of the resulting solution depends on the concentration of H+ ions present.
Option a: Fe3+(aq) + 3H2O(l) <--> Fe(OH)3(aq) + 3H+(aq)
This equation represents the hydrolysis of Fe3+ ions, which can contribute to the generation of H+ ions in the solution. However, Fe(NO3)3 does not readily undergo hydrolysis to a significant extent.
Option b: NO3-(aq) + H2O(l) <--> HNO3(aq) + OH-(aq)
This equation represents the dissociation of nitrate ions (NO3-) into hydronium (H3O+) and hydroxide (OH-) ions. However, this dissociation does not account for the observed pH deviation in the given Fe(NO3)3 solution.
Option c: Fe(H2O)63+(aq) + H2O(l) <--> Fe(H2O)5(OH)2+(aq) + H3O+(aq)
This equation represents the hydrolysis of the hydrated Fe3+ ions, resulting in the formation of H3O+ ions, which contribute to the observed pH deviation. Therefore, this option explains the observation accurately.
Option d: Fe(H2O)63+(aq) + H2O(l) <--> Fe(OH2)5(H3O)4+(aq) + OH-(aq)
This equation represents the hydrolysis of the hydrated Fe3+ ions with the formation of hydroxide ions (OH-) and hydronium ions (H3O+). However, this hydrolysis reaction is less likely to occur compared to the hydrolysis reaction described in option c.
Option e: HNO3(aq) + H2O(l) <--> H3O+(aq) + NO3-(aq)
This equation represents the dissociation of nitric acid (HNO3) into hydronium (H3O+) and nitrate (NO3-) ions. While HNO3 can contribute to H+ ions in solution, it does not fully address the pH deviation in the Fe(NO3)3 solution.
Therefore, option c is the best equation that accounts for the observed pH deviation in the given Fe(NO3)3 solution.
2. For question 2, you correctly calculated the concentration of H+ ions by using the equation 10^(-pH). However, to find the percentage ionization of NH3 in the solution, you need to compare the concentration of the ionized form (NH4+) to the initial concentration of NH3.
NH3(aq) + H2O(l) <--> NH4+(aq) + OH-(aq)
The dissociation of NH3 in water results in the formation of NH4+ ions and hydroxide (OH-) ions. Since NH4+ is the ionized form, you need to determine the concentration of NH4+.
Since NH3 is a weak base, only a fraction of it will ionize in water. The equilibrium expression for the dissociation is represented as follows:
[NH4+] / [NH3] = Ka
To simplify the calculation, we will assume that the concentration of OH- ions is negligible compared to the concentration of NH4+. Therefore, the [OH-] can be ignored. The concentration of NH4+ can be calculated by subtracting the [H+] concentration from the initial concentration of NH3.
[H+] = 10^(-pH)
[NH4+] = [NH3] - [H+]
Substituting the given values, [NH3] = 0.05 mol/L and [H+] = 2 x 10^(-8) mol/L, we can calculate [NH4+].
[NH4+] = (0.05 mol/L) - (2 x 10^(-8) mol/L)
Finally, to find the percentage ionization, we divide the concentration of NH4+ by the initial concentration of NH3 and multiply by 100%.
Percentage ionization = ([NH4+] / [NH3]) x 100%
Plugging in the calculated values, we find:
Percentage ionization = ( [NH4+] / [NH3] ) x 100%
Percentage ionization = ( (0.05 mol/L) - (2 x 10^(-8) mol/L) / (0.05 mol/L ) ) x 100%
Percentage ionization ≈ 3%
Therefore, option c (3%) is the correct answer for the percentage ionization in the given NH3 solution with a pH of 11.00.