Solubility of a tribasic weak acid
$begingroup$
I have a question that reads:
$ce{H3A}$ is a tribasic acid. The three deprotonations can be written as
$ce{H3A <=>[pKa1 = 2.9] H2A- <=>[pKa2 = 4.2] HA^{2-} <=>[pKa3 = 5.2] A^{3-}}$
What is the predominant species in a 0.10M solution of $ce{H3A}$ at pH 5?
a) $ce{H3A}$
b) $ce{H2A-}$
c) $ce{HA^{2-}}$
d) $ce{A^{3-}}$
I worked this out as
$ce{Ka1 = frac{[H+][H2A-]}{[H3A]}}, ce{Ka2 = frac{[H+][HA^{2-}]}{[H2A-]}}, ce{Ka3 = frac{[H+][A^{3-}]}{[HA^{2-}]}}$
At pH 5, $ce{[H+] = 10^{-5}M}$
So,
$ce{[H+] = 10^{-5} = Ka1frac{[H2A-]}{[H3A]} = 10^{-2.9} frac{[H2A-]}{[H3A]}}ce{implies frac{[H2A-]}{[H3A]} = 10^{2.1} }$
and similarly,
$ce{frac{[HA2-]}{[H2A-]} = 10^{0.8}}$
$ce{frac{[A^{3-}]}{[HA^{2-}]} = 10^{-0.2}}$
Since $ce{[HA^{2-}] > [H3A], [HA^{2-}]>[H2A-] and [A^{3-}]<[HA^{2-}]}$, the predominant species is $ce{HA^{2-}}$, so c).
This is the correct answer according to the solutions, but I'm not sure if my logic is correct, or if there's an easier way to arrive at the same conclusion (it's significant working out for a MCQ).
solubility
asked 13 hours ago
GremlinGremlin
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$begingroup$
I have a question that reads:
$ce{H3A}$ is a tribasic acid. The three deprotonations can be written as
$ce{H3A <=>[pKa1 = 2.9] H2A- <=>[pKa2 = 4.2] HA^{2-} <=>[pKa3 = 5.2] A^{3-}}$
What is the predominant species in a 0.10M solution of $ce{H3A}$ at pH 5?
a) $ce{H3A}$
b) $ce{H2A-}$
c) $ce{HA^{2-}}$
d) $ce{A^{3-}}$
I worked this out as
$ce{Ka1 = frac{[H+][H2A-]}{[H3A]}}, ce{Ka2 = frac{[H+][HA^{2-}]}{[H2A-]}}, ce{Ka3 = frac{[H+][A^{3-}]}{[HA^{2-}]}}$
At pH 5, $ce{[H+] = 10^{-5}M}$
So,
$ce{[H+] = 10^{-5} = Ka1frac{[H2A-]}{[H3A]} = 10^{-2.9} frac{[H2A-]}{[H3A]}}ce{implies frac{[H2A-]}{[H3A]} = 10^{2.1} }$
and similarly,
$ce{frac{[HA2-]}{[H2A-]} = 10^{0.8}}$
$ce{frac{[A^{3-}]}{[HA^{2-}]} = 10^{-0.2}}$
Since $ce{[HA^{2-}] > [H3A], [HA^{2-}]>[H2A-] and [A^{3-}]<[HA^{2-}]}$, the predominant species is $ce{HA^{2-}}$, so c).
This is the correct answer according to the solutions, but I'm not sure if my logic is correct, or if there's an easier way to arrive at the same conclusion (it's significant working out for a MCQ).
solubility
asked 13 hours ago
GremlinGremlin
1236
New contributor
Gremlin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have a question that reads:
$ce{H3A}$ is a tribasic acid. The three deprotonations can be written as
$ce{H3A <=>[pKa1 = 2.9] H2A- <=>[pKa2 = 4.2] HA^{2-} <=>[pKa3 = 5.2] A^{3-}}$
What is the predominant species in a 0.10M solution of $ce{H3A}$ at pH 5?
a) $ce{H3A}$
b) $ce{H2A-}$
c) $ce{HA^{2-}}$
d) $ce{A^{3-}}$
I worked this out as
$ce{Ka1 = frac{[H+][H2A-]}{[H3A]}}, ce{Ka2 = frac{[H+][HA^{2-}]}{[H2A-]}}, ce{Ka3 = frac{[H+][A^{3-}]}{[HA^{2-}]}}$
At pH 5, $ce{[H+] = 10^{-5}M}$
So,
$ce{[H+] = 10^{-5} = Ka1frac{[H2A-]}{[H3A]} = 10^{-2.9} frac{[H2A-]}{[H3A]}}ce{implies frac{[H2A-]}{[H3A]} = 10^{2.1} }$
and similarly,
$ce{frac{[HA2-]}{[H2A-]} = 10^{0.8}}$
$ce{frac{[A^{3-}]}{[HA^{2-}]} = 10^{-0.2}}$
Since $ce{[HA^{2-}] > [H3A], [HA^{2-}]>[H2A-] and [A^{3-}]<[HA^{2-}]}$, the predominant species is $ce{HA^{2-}}$, so c).
This is the correct answer according to the solutions, but I'm not sure if my logic is correct, or if there's an easier way to arrive at the same conclusion (it's significant working out for a MCQ).
solubility
asked 13 hours ago
GremlinGremlin
1236
New contributor
Gremlin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have a question that reads:
$ce{H3A}$ is a tribasic acid. The three deprotonations can be written as
$ce{H3A <=>[pKa1 = 2.9] H2A- <=>[pKa2 = 4.2] HA^{2-} <=>[pKa3 = 5.2] A^{3-}}$
What is the predominant species in a 0.10M solution of $ce{H3A}$ at pH 5?
a) $ce{H3A}$
b) $ce{H2A-}$
c) $ce{HA^{2-}}$
d) $ce{A^{3-}}$
I worked this out as
$ce{Ka1 = frac{[H+][H2A-]}{[H3A]}}, ce{Ka2 = frac{[H+][HA^{2-}]}{[H2A-]}}, ce{Ka3 = frac{[H+][A^{3-}]}{[HA^{2-}]}}$
At pH 5, $ce{[H+] = 10^{-5}M}$
So,
$ce{[H+] = 10^{-5} = Ka1frac{[H2A-]}{[H3A]} = 10^{-2.9} frac{[H2A-]}{[H3A]}}ce{implies frac{[H2A-]}{[H3A]} = 10^{2.1} }$
and similarly,
$ce{frac{[HA2-]}{[H2A-]} = 10^{0.8}}$
$ce{frac{[A^{3-}]}{[HA^{2-}]} = 10^{-0.2}}$
Since $ce{[HA^{2-}] > [H3A], [HA^{2-}]>[H2A-] and [A^{3-}]<[HA^{2-}]}$, the predominant species is $ce{HA^{2-}}$, so c).
This is the correct answer according to the solutions, but I'm not sure if my logic is correct, or if there's an easier way to arrive at the same conclusion (it's significant working out for a MCQ).
solubility
solubility
asked 13 hours ago
GremlinGremlin
1236
New contributor
Gremlin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 13 hours ago
GremlinGremlin
1236
New contributor
Gremlin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 12 hours ago
Gremlin
asked 13 hours ago
GremlinGremlin
1236
New contributor
Gremlin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 13 hours ago
GremlinGremlin
1236
asked 13 hours ago
GremlinGremlin
1236
1236
New contributor
Gremlin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Gremlin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
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1 Answer
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$begingroup$
You can also use the Henderson-Hasselbalch equation:
$$mathrm{pH} = mathrm{p}K_{mathrm{a}} + log frac{ce{[A-]}}{ce{[HA]}}$$
rearranges to:
$$mathrm{pH} - mathrm{p}K_{mathrm{a}} = log frac{ce{[A-]}}{ce{[HA]}}$$
$log frac{ce{[A-]}}{ce{[HA]}} > 0$ for conjugate base dominating.
$log frac{ce{[A-]}}{ce{[HA]}} < 0$ for acid dominating.
$log frac{ce{[A-]}}{ce{[HA]}} = 0$ if they're present in the same concentrations.
The equilibrium of interest is the third dissociation.
Here, the LHS equaluates to $-0.2$ indicating that the acid in that equation ($ce{HA^{2-}}$) dominates over the conjugate base ($ce{A^{3-}}$). You should be able to extend this for each equilibrium.
answered 13 hours ago
ZheZhe
12.7k12550
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1 Answer
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$begingroup$
You can also use the Henderson-Hasselbalch equation:
$$mathrm{pH} = mathrm{p}K_{mathrm{a}} + log frac{ce{[A-]}}{ce{[HA]}}$$
rearranges to:
$$mathrm{pH} - mathrm{p}K_{mathrm{a}} = log frac{ce{[A-]}}{ce{[HA]}}$$
$log frac{ce{[A-]}}{ce{[HA]}} > 0$ for conjugate base dominating.
$log frac{ce{[A-]}}{ce{[HA]}} < 0$ for acid dominating.
$log frac{ce{[A-]}}{ce{[HA]}} = 0$ if they're present in the same concentrations.
The equilibrium of interest is the third dissociation.
Here, the LHS equaluates to $-0.2$ indicating that the acid in that equation ($ce{HA^{2-}}$) dominates over the conjugate base ($ce{A^{3-}}$). You should be able to extend this for each equilibrium.
answered 13 hours ago
ZheZhe
12.7k12550
$endgroup$
add a comment |
$begingroup$
You can also use the Henderson-Hasselbalch equation:
$$mathrm{pH} = mathrm{p}K_{mathrm{a}} + log frac{ce{[A-]}}{ce{[HA]}}$$
rearranges to:
$$mathrm{pH} - mathrm{p}K_{mathrm{a}} = log frac{ce{[A-]}}{ce{[HA]}}$$
$log frac{ce{[A-]}}{ce{[HA]}} > 0$ for conjugate base dominating.
$log frac{ce{[A-]}}{ce{[HA]}} < 0$ for acid dominating.
$log frac{ce{[A-]}}{ce{[HA]}} = 0$ if they're present in the same concentrations.
The equilibrium of interest is the third dissociation.
Here, the LHS equaluates to $-0.2$ indicating that the acid in that equation ($ce{HA^{2-}}$) dominates over the conjugate base ($ce{A^{3-}}$). You should be able to extend this for each equilibrium.
answered 13 hours ago
ZheZhe
12.7k12550
$endgroup$
add a comment |
$begingroup$
You can also use the Henderson-Hasselbalch equation:
$$mathrm{pH} = mathrm{p}K_{mathrm{a}} + log frac{ce{[A-]}}{ce{[HA]}}$$
rearranges to:
$$mathrm{pH} - mathrm{p}K_{mathrm{a}} = log frac{ce{[A-]}}{ce{[HA]}}$$
$log frac{ce{[A-]}}{ce{[HA]}} > 0$ for conjugate base dominating.
$log frac{ce{[A-]}}{ce{[HA]}} < 0$ for acid dominating.
$log frac{ce{[A-]}}{ce{[HA]}} = 0$ if they're present in the same concentrations.
The equilibrium of interest is the third dissociation.
Here, the LHS equaluates to $-0.2$ indicating that the acid in that equation ($ce{HA^{2-}}$) dominates over the conjugate base ($ce{A^{3-}}$). You should be able to extend this for each equilibrium.
answered 13 hours ago
ZheZhe
12.7k12550
$endgroup$
You can also use the Henderson-Hasselbalch equation:
$$mathrm{pH} = mathrm{p}K_{mathrm{a}} + log frac{ce{[A-]}}{ce{[HA]}}$$
rearranges to:
$$mathrm{pH} - mathrm{p}K_{mathrm{a}} = log frac{ce{[A-]}}{ce{[HA]}}$$
$log frac{ce{[A-]}}{ce{[HA]}} > 0$ for conjugate base dominating.
$log frac{ce{[A-]}}{ce{[HA]}} < 0$ for acid dominating.
$log frac{ce{[A-]}}{ce{[HA]}} = 0$ if they're present in the same concentrations.
The equilibrium of interest is the third dissociation.
Here, the LHS equaluates to $-0.2$ indicating that the acid in that equation ($ce{HA^{2-}}$) dominates over the conjugate base ($ce{A^{3-}}$). You should be able to extend this for each equilibrium.
answered 13 hours ago
ZheZhe
12.7k12550
answered 13 hours ago
ZheZhe
12.7k12550
answered 13 hours ago
ZheZhe
12.7k12550
answered 13 hours ago
ZheZhe
12.7k12550
12.7k12550
add a comment |
add a comment |
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