NH3(aq)+H2O(l)⇄NH4+(aq)+OH−(aq)
Kb=1.8×10−5
at 25C
Initial [NH3] | Equilibrium [NH4+] | Equilibrium [OH−] | pOH
0.15 | 1.6×10−3 | 1.6×10−3 | 2.78
0.30 | 2.3×10−3 | ? | ?
NH3 is a weak base that reacts with water according to the chemical equilibrium represented above. The table provides some information for two NH3(aq) solutions of different concentration at 25°C . Which of the following is true about the more concentrated 0.30MNH3(aq) , and why?
A. [OH−]=3.2×10−3M and pOH<2.78
B. [OH−]=3.2×10−3M and pOH>2.78
C. [OH−]=2.3×10−3M and pOH<2.78
D. [OH−]=2.3×10−3M and pOH<2.78
Why?
A. because a higher [OH−] corresponds to a lower pOH for 0.30MNH3(aq) compared to 0.15MNH3(aq) .
B. because a higher [OH−] corresponds to a higher pOH for 0.30MNH3(aq) compared to 0.15MNH3(aq) .
NH3(aq)+H2O(l)⇄NH4+(aq)+OH−(aq)..........Kb=1.8×10−5
Your table is hard to read. I think I can help that.
................. [NH3]............... [NH4+] ........[OH−]........ pOH
Initial.........0.15 ................ 1.6×10−3 ... 1.6×10−3 ..... 2.78
Initial.........0.30................. 2.3×10−3 ......2.3x10^-3 .....2.63
I see that both c and d or why are the same AND I confused about what you want. I've calculated the question marks you have so you can make the decision for the answer.
A. OH−]=3.2×10−3M and pOH<2.78 , because a higher [OH−] corresponds to a lower pOH for 0.30MNH3(aq) compared to 0.15MNH3(aq) .
B. [OH−]=3.2×10−3M and pOH>2.78, because a higher [OH−] corresponds to a higher pOH for 0.30MNH3(aq) compared to 0.15MNH3(aq).
C. [OH−]=2.3×10−3M and pOH<2.78, because a higher [OH−] corresponds to a lower pOH for 0.30MNH3(aq) compared to 0.15MNH3(aq).
D. [OH−]=2.3×10−3M and pOH>2.78, because a higher [OH−] corresponds to a higher pOH for 0.30MNH3(aq) compared to 0.15MNH3(aq).
Well, let's get to the bottom of this! We know that a higher concentration of NH3 corresponds to a higher concentration of NH4+ at equilibrium. And since NH4+ and OH- are produced in a 1:1 ratio, a higher concentration of NH4+ means a higher concentration of OH-. So, that eliminates options C and D.
Now, looking at the pOH values, we see that the pOH of the 0.15MNH3(aq) solution is 2.78. If the concentration of OH- in the 0.30MNH3(aq) solution is 3.2×10−3M, then the pOH would actually be lower than 2.78.
So, the correct answer is A. A higher [OH-] corresponds to a lower pOH for 0.30MNH3(aq) compared to 0.15MNH3(aq). It seems like the more concentrated solution is playing by its own rules and clowning around with the pOH values!
The correct answer is B. [OH−]=3.2×10−3M and pOH>2.78.
Explanation:
According to the information given in the table, for the 0.15M NH3(aq) solution, the equilibrium [OH−] is 1.6×10−3M, and the pOH is 2.78. For the 0.30M NH3(aq) solution, the equilibrium [NH4+] is 2.3×10−3M, but the equilibrium [OH−] is not given.
Kb is the equilibrium constant for this reaction, which is a measure of the strength of a base. The larger the value of Kb, the stronger the base. In this case, Kb is given as 1.8×10−5.
Since the concentration of NH3 has doubled from 0.15M to 0.30M, we can expect the concentration of NH4+ and OH− to increase as well. This is because the forward reaction will be favored to establish a new equilibrium.
Because Kb is small, NH3 is a weak base and will not completely ionize in water. As a result, some NH3 will remain and not fully react with water to form NH4+ and OH−. Therefore, we know that the concentration of OH− in the 0.30M NH3(aq) solution will be higher than 1.6×10−3M, which is the equilibrium OH− concentration for the 0.15M NH3(aq) solution.
Additionally, since OH− concentration increased, pOH will also increase. A higher pOH corresponds to a lower pH. Therefore, the pOH for the 0.30M NH3(aq) solution will be greater than 2.78.
Therefore, the correct answer is B. [OH−]=3.2×10−3M and pOH>2.78.
To determine the correct answer, we need to analyze the values provided in the table and apply our understanding of the equilibrium equation and the concept of pOH.
The equilibrium equation shows that NH3 reacts with water to produce NH4+ and OH-. It is a weak base with a Kb value of 1.8x10^-5.
By comparing the equilibrium concentrations of NH4+ and OH- for the two NH3 solutions, we can determine how the concentration of NH3 affects the concentrations of NH4+ and OH-.
For the 0.15M NH3 solution:
- The equilibrium [NH4+] is 1.6x10^-3M.
- The equilibrium [OH-] is 1.6x10^-3M.
For the 0.30M NH3 solution:
- We are given the equilibrium [NH4+], which is 2.3x10^-3M.
- We need to find the equilibrium [OH-] and calculate the corresponding pOH.
To find the equilibrium [OH-], we can use the Kb expression:
Kb = [NH4+][OH-] / [NH3]
We know the Kb value is 1.8x10^-5 and the equilibrium [NH4+] is 2.3x10^-3M. The equilibrium [NH3] for the 0.30M NH3 solution is 0.30M. Substituting these values into the equation, we can solve for [OH-]:
1.8x10^-5 = (2.3x10^-3M)([OH-]) / 0.30M
Cross-multiplying and rearranging the equation, we get:
[OH-] = (1.8x10^-5)(0.30M) / (2.3x10^-3M)
Using a calculator, we find that [OH-] is approximately 2.383x10^-4M.
Now, we can calculate the pOH:
pOH = -log[OH-] = -log(2.383x10^-4) ≈ 3.62
Comparing the pOH values obtained for the two NH3 solutions, we can see that pOH for the 0.30M NH3 solution is greater than 2.78, which was the pOH value for the 0.15M NH3 solution.
Therefore, the correct answer is B. [OH-] = 3.2x10^-3M and pOH > 2.78, as a higher [OH-] corresponds to a higher pOH for the 0.30M NH3 solution compared to the 0.15M NH3 solution.