Sapling week 2 #8
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 113
- Joined: Wed Sep 30, 2020 10:05 pm
Sapling week 2 #8
I'm a little confused on this question. It says "H3 is a weak base ( Kb=1.8×10−5 ) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.019 M in NH4Cl at 25 °C?"
Can anyone walk me through how to solve for this problem?
Can anyone walk me through how to solve for this problem?
-
- Posts: 18
- Joined: Thu Jul 25, 2019 12:16 am
Re: Sapling week 2 #8
Hi Lexy!
This personally how I went about solving this problem:
1. Since Kb is given to you in this problem and the constant for Kw is known as 1.0 x 10^-14, I solved for Ka using the equation: Ka x Kb = Kw.
2. Once you calculate the value for Ka, you should notice that it is significantly small; this means that we can use the approximation rule to solve for X.
3. But before doing that, we must first create an ICE table using the initial concentration given to you by the problem in order to determine some of the values/ expressions that will be plugged into your equilibrium constant expression for Ka.
4. After doing so, you can solve for X using the approximation rule that I mentioned earlier. (Remember that X=[H3O+]; if this doesn't make sense to you, refer back to your ICE table).
5. Now that you have solved for X aka [H3O+], you can use the equation pH=-log[H3O+] to solve for the pH!
and 6. If you would like to verify that the assumption made earlier for X was valid, you can now use the 5% rule as well to check. This is done by diving the value that you found for X over the initial concentration given to you for NH4+ and multiplying that by 100%!
I hope this helped; let me know if you have any questions!
This personally how I went about solving this problem:
1. Since Kb is given to you in this problem and the constant for Kw is known as 1.0 x 10^-14, I solved for Ka using the equation: Ka x Kb = Kw.
2. Once you calculate the value for Ka, you should notice that it is significantly small; this means that we can use the approximation rule to solve for X.
3. But before doing that, we must first create an ICE table using the initial concentration given to you by the problem in order to determine some of the values/ expressions that will be plugged into your equilibrium constant expression for Ka.
4. After doing so, you can solve for X using the approximation rule that I mentioned earlier. (Remember that X=[H3O+]; if this doesn't make sense to you, refer back to your ICE table).
5. Now that you have solved for X aka [H3O+], you can use the equation pH=-log[H3O+] to solve for the pH!
and 6. If you would like to verify that the assumption made earlier for X was valid, you can now use the 5% rule as well to check. This is done by diving the value that you found for X over the initial concentration given to you for NH4+ and multiplying that by 100%!
I hope this helped; let me know if you have any questions!
-
- Posts: 108
- Joined: Wed Sep 30, 2020 9:48 pm
- Been upvoted: 1 time
Re: Sapling week 2 #8
Because you are given the concentration of solution in Nh4Cl, which you are told acts as a weak acid, you are going to want to use the Ka to find the concentration of H3O+... so use kw to convert the given kb to ka, then do the ICE table, etc. like the previous SAPLING problems (rather than try to use the given Kb to find [OH-] and pOH to find pH, since this may seem like a logical route with the problem starting off with the mention of NH3 being a weak base)
-
- Posts: 52
- Joined: Wed Dec 02, 2020 12:19 am
Re: Sapling week 2 #8
What helps is actually starting to write the reaction down before you actually start doing anything. So you have the products and reactants right there in front of you. NH3 and H3O and NH4Cl. So the ice box making becomes easier.
-
- Posts: 100
- Joined: Wed Sep 30, 2020 9:46 pm
Re: Sapling week 2 #8
Saja Kamal 2L wrote:Hi Lexy!
This personally how I went about solving this problem:
1. Since Kb is given to you in this problem and the constant for Kw is known as 1.0 x 10^-14, I solved for Ka using the equation: Ka x Kb = Kw.
2. Once you calculate the value for Ka, you should notice that it is significantly small; this means that we can use the approximation rule to solve for X.
3. But before doing that, we must first create an ICE table using the initial concentration given to you by the problem in order to determine some of the values/ expressions that will be plugged into your equilibrium constant expression for Ka.
4. After doing so, you can solve for X using the approximation rule that I mentioned earlier. (Remember that X=[H3O+]; if this doesn't make sense to you, refer back to your ICE table).
5. Now that you have solved for X aka [H3O+], you can use the equation pH=-log[H3O+] to solve for the pH!
and 6. If you would like to verify that the assumption made earlier for X was valid, you can now use the 5% rule as well to check. This is done by diving the value that you found for X over the initial concentration given to you for NH4+ and multiplying that by 100%!
I hope this helped; let me know if you have any questions!
so, because it says "NH_3 is a weak base" we know to change the given Ka to Kb instead?
-
- Posts: 111
- Joined: Wed Sep 30, 2020 9:45 pm
- Been upvoted: 2 times
Re: Sapling week 2 #8
Kandyce Lance 3E wrote:Saja Kamal 2L wrote:Hi Lexy!
This personally how I went about solving this problem:
1. Since Kb is given to you in this problem and the constant for Kw is known as 1.0 x 10^-14, I solved for Ka using the equation: Ka x Kb = Kw.
2. Once you calculate the value for Ka, you should notice that it is significantly small; this means that we can use the approximation rule to solve for X.
3. But before doing that, we must first create an ICE table using the initial concentration given to you by the problem in order to determine some of the values/ expressions that will be plugged into your equilibrium constant expression for Ka.
4. After doing so, you can solve for X using the approximation rule that I mentioned earlier. (Remember that X=[H3O+]; if this doesn't make sense to you, refer back to your ICE table).
5. Now that you have solved for X aka [H3O+], you can use the equation pH=-log[H3O+] to solve for the pH!
and 6. If you would like to verify that the assumption made earlier for X was valid, you can now use the 5% rule as well to check. This is done by diving the value that you found for X over the initial concentration given to you for NH4+ and multiplying that by 100%!
I hope this helped; let me know if you have any questions!
so, because it says "NH_3 is a weak base" we know to change the given Ka to Kb instead?
We are given Kb and need to convert to Ka. We know to do this because it says: "the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.019 M in NH4Cl at 25 °C?"
Since it is asking us to find something using NH4Cl, we must use Ka since NH4CL is an acid. As we are given Kb, we must change it to Ka.
-
- Posts: 124
- Joined: Wed Sep 30, 2020 9:43 pm
Re: Sapling week 2 #8
This whole thread was amazing! I had so much trouble with this problem and ts all helped so much!
-
- Posts: 101
- Joined: Wed Sep 30, 2020 9:35 pm
Re: Sapling week 2 #8
I first wrote the chemical reaction being NH4+ + H20 yields NH3 + H3O+. From there I make an ice chart and put the given M of NH4Cl for my initial concentration of NH4+. I then find the ka value by using the equation ka x kb = 10^-14. Set up an equation for ka and then use negative log to find the pH.
-
- Posts: 101
- Joined: Wed Sep 30, 2020 9:42 pm
Re: Sapling week 2 #8
Just for clarification, one of the first steps is to calculate Ka because we know both Kw and Kb, so we can solve for the third value in the formula (Ka = Kw / Kb). We then use the Ka to find the concentration in terms of the NH4Cl.
-
- Posts: 100
- Joined: Wed Sep 30, 2020 9:46 pm
Re: Sapling week 2 #8
Amanda Bueno-Kling wrote:Kandyce Lance 3E wrote:Saja Kamal 2L wrote:Hi Lexy!
This personally how I went about solving this problem:
1. Since Kb is given to you in this problem and the constant for Kw is known as 1.0 x 10^-14, I solved for Ka using the equation: Ka x Kb = Kw.
2. Once you calculate the value for Ka, you should notice that it is significantly small; this means that we can use the approximation rule to solve for X.
3. But before doing that, we must first create an ICE table using the initial concentration given to you by the problem in order to determine some of the values/ expressions that will be plugged into your equilibrium constant expression for Ka.
4. After doing so, you can solve for X using the approximation rule that I mentioned earlier. (Remember that X=[H3O+]; if this doesn't make sense to you, refer back to your ICE table).
5. Now that you have solved for X aka [H3O+], you can use the equation pH=-log[H3O+] to solve for the pH!
and 6. If you would like to verify that the assumption made earlier for X was valid, you can now use the 5% rule as well to check. This is done by diving the value that you found for X over the initial concentration given to you for NH4+ and multiplying that by 100%!
I hope this helped; let me know if you have any questions!
so, because it says "NH_3 is a weak base" we know to change the given Ka to Kb instead?
We are given Kb and need to convert to Ka. We know to do this because it says: "the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.019 M in NH4Cl at 25 °C?"
Since it is asking us to find something using NH4Cl, we must use Ka since NH4CL is an acid. As we are given Kb, we must change it to Ka.
ohh ok that makes sense thank you!!
-
- Posts: 111
- Joined: Wed Sep 30, 2020 9:45 pm
- Been upvoted: 2 times
Re: Sapling week 2 #8
Kandyce Lance 3E wrote:Amanda Bueno-Kling wrote:Kandyce Lance 3E wrote:so, because it says "NH_3 is a weak base" we know to change the given Ka to Kb instead?
We are given Kb and need to convert to Ka. We know to do this because it says: "the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.019 M in NH4Cl at 25 °C?"
Since it is asking us to find something using NH4Cl, we must use Ka since NH4CL is an acid. As we are given Kb, we must change it to Ka.
ohh ok that makes sense thank you!!
Glad I could help you!
Return to “Non-Equilibrium Conditions & The Reaction Quotient”
Who is online
Users browsing this forum: No registered users and 9 guests