Can someone please walk me through this problem?
The Kb for an amine is 1.339×10−5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.870 ? Assume that all OH− came from the reaction of B with H2O.
Week 2 Achieve HW Question 5
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Re: Week 2 Achieve HW Question 5
Here was the method I used:
1. Use the pH value to find the pOH value since you are working with OH-
2. take the negative log of pOH to find [OH-] at equilibrium
3. use the given Kb value and solve the quadratic
4. calculate the initial concentration
5. divide the equilibrium concentration by the initial and multiply by 100 in order to get the percent protonated
Hope this helps!
1. Use the pH value to find the pOH value since you are working with OH-
2. take the negative log of pOH to find [OH-] at equilibrium
3. use the given Kb value and solve the quadratic
4. calculate the initial concentration
5. divide the equilibrium concentration by the initial and multiply by 100 in order to get the percent protonated
Hope this helps!
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Re: Week 2 Achieve HW Question 5
First we'd start off with the main chemical equation being presented in the problem:
B(aq) + H2O(l) <=> BH+(aq) + OH-(aq)
From the equation we can see that there is one mole for both BH+ and OH- meaning that BH+ = OH-.
Using the basicity constant (below) we would be able to find the formal concentration of B, which represents the amine. (Reaarange the formula to find B)
Kb = [BH+][OH-] / [B]
To find the concentration of OH- we would need to determine the pOH from the given pH. To do this we would have to subtract 14 by the pH because we know that pH + pOH = 14.
14 - pH = ?
With the value from 14 - pH, we would use the inverse log and get our concentration of OH- and BH+.
10^( - ?)
Plug in the value to the basicity constant and solve for B.
To find the percentage of the protonated amine, you would have to divide the concentration of BH+ by the formal concentration of B, which is the sum of BH+ and B, finally multiplying by 100 to get the percentage
([BH+] / [BH+] + [B]) * 100 = % of protonated amine
B(aq) + H2O(l) <=> BH+(aq) + OH-(aq)
From the equation we can see that there is one mole for both BH+ and OH- meaning that BH+ = OH-.
Using the basicity constant (below) we would be able to find the formal concentration of B, which represents the amine. (Reaarange the formula to find B)
Kb = [BH+][OH-] / [B]
To find the concentration of OH- we would need to determine the pOH from the given pH. To do this we would have to subtract 14 by the pH because we know that pH + pOH = 14.
14 - pH = ?
With the value from 14 - pH, we would use the inverse log and get our concentration of OH- and BH+.
10^( - ?)
Plug in the value to the basicity constant and solve for B.
To find the percentage of the protonated amine, you would have to divide the concentration of BH+ by the formal concentration of B, which is the sum of BH+ and B, finally multiplying by 100 to get the percentage
([BH+] / [BH+] + [B]) * 100 = % of protonated amine
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Re: Week 2 Achieve HW Question 5
Hello,
First you want to use the pH value given in the problem to solve for pOH.
Then you want to use your calculated pOH to find the concentration of both OH and BH since they will be equal in a Kb equation.
then you use your calculated concentrations and plug it into the Kb equation with the given Kb value in order to solve for the concentration of B
Lastly you solve for percent protonated by doing [BH]/[Bformal]
First you want to use the pH value given in the problem to solve for pOH.
Then you want to use your calculated pOH to find the concentration of both OH and BH since they will be equal in a Kb equation.
then you use your calculated concentrations and plug it into the Kb equation with the given Kb value in order to solve for the concentration of B
Lastly you solve for percent protonated by doing [BH]/[Bformal]
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