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Solubility of a tribasic weak acid

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4

$begingroup$

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

share|improve this question

edited 14 hours ago

Gremlin

asked 14 hours ago

GremlinGremlin

1236

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4

$begingroup$

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

share|improve this question

edited 14 hours ago

Gremlin

asked 14 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

share|improve this question

edited 14 hours ago

Gremlin

asked 14 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.

$endgroup$

share|improve this question

edited 14 hours ago

Gremlin

asked 14 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.

share|improve this question

edited 14 hours ago

Gremlin

asked 14 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 14 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 14 hours ago

GremlinGremlin

1236

1 Answer
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5

$begingroup$

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.

share|improve this answer

answered 14 hours ago

ZheZhe

12.7k12550

$endgroup$

add a comment | 

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5

$begingroup$

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.

share|improve this answer

answered 14 hours ago

ZheZhe

12.7k12550

$endgroup$

add a comment | 

5

$begingroup$

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.

share|improve this answer

answered 14 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.

share|improve this answer

answered 14 hours ago

ZheZhe

12.7k12550

$endgroup$

share|improve this answer

answered 14 hours ago

ZheZhe

12.7k12550

answered 14 hours ago

ZheZhe

12.7k12550

answered 14 hours ago

ZheZhe

12.7k12550