What Is A Monoprotic Acid

Polyprotic Acids & Bases

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Co-ordinate to Brønsted and Lowry an acid is a proton donor and a base is a proton acceptor. This idea of proton donor and proton acceptor is important in agreement monoprotic and polyprotic acids and bases because monoprotic corresponds to the transfer of one proton and polyprotic refers to the transfer of more than i proton. Therefore, a monoprotic acid is an acid that tin can donate only i proton, while polyprotic acrid tin donate more than 1 proton. Similarly, a mon oprotic base can but accept one proton, while a polyprotic base can accept more than ane proton.

Introduction

One mode to display the differences betwixt monoprotic and polyprotic acids and bases is through titration, which clearly depicts the equivalence points and acid or base dissociation constants. The acid dissociation abiding, signified past \(K_a\), and the base dissociation abiding, \(K_b\), are equilibrium constants for the dissociation of weak acids and weak bases. The larger the value of either \(K_a\) or \(K_b\) signifies a stronger acid or base, respectively.

Here is a list of of import equations and constants when dealing with \(K_a\) and \(K_b\):

For the general equation of a weak acid,

\[HA_{(aq)} + H_2O_{(l)} \Longleftrightarrow H_3O^+_{(aq)} + A^-_{(aq)} \label{ane}\]

you demand to solve for the \(K_a\) value. To do that you use

\[K_a = \dfrac{[H_3O^+][A^-]}{[HA]} \label{two}\]

Another necessary value is the \(pK_a\) value, and that is obtained through \(pK_a = {-logK_a}\)

The procedure is very like for weak bases. The general equation of a weak base is

\[BOH \Longleftrightarrow B^+ + OH^- \label{3}\]

Solving for the \(K_b\)value is the same as the \(K_a\) value. Yous use the formula

\[K_b = \dfrac{[B^+][OH^-]}{[BOH]} \characterization{4}\]

The \(pK_b\) value is found through \(pK_b = {-logK_b}\)

The \(K_w\) value is found with\(K_w = {[H3O^+]}{[OH^-]}\).

\[K_w = 1.0 \times 10^{-14} \label{five}\]

Monoprotic Acids

Monoprotic acids are acids that can release only one proton per molecule and take one equivalence point.

Here is a table of some mutual monoprotic acids:

Name	Formula	\(\pmb{K_a}\)
Muriatic acid (stiff)	HCl	ane.3 ten 10⁶
Nitric acrid (strong)	HNO₃	two.4 ten x^ane
Acetic acid (weak)	CH_iiiCOOH	1.74 x 10^-five

Monoprotic Bases

Monoprotic Bases are bases that can merely react with ane proton per molecule and similar to monoprotic acids, just have 1 equivalence point. Hither is a list of some common monoprotic bases:

Proper name	Formula	\(\pmb{K_b}\)
Sodium hydroxide (strong)	NaOH	six.iii X 10^-ane
Potassium hydroxide (strong)	KOH	3.xvi Ten 10^-ane
Ammonia (weak)	NH₃	1.80 x x^-5

Example \(\PageIndex{one}\)

What is the pH of the solution that results from the improver of 200 mL of 0.ane Thousand CsOH(aq) to 50 mL of 0.2M HNO₂(aq)? (pKa= 3.xiv for HNO₂)

Solution

\[\dfrac{0.1 mol}{Fifty}*200 mL* \dfrac{1 50}{chiliad mL} = {0.02 mol CsOH}\]

\[\dfrac{0.ii mol}{L}*50 mL* \dfrac{ane 50}{yard mL} = {0.01 mol HNO_2}\]

Then do an Ice Table for

\[CsOH + HNO_2 \Longleftrightarrow H_2O + CsNO_2\]

yielding \([CsOH]= [OH^-]= 0.01M\)

Then to find pH first we observe pOH \(pOH = {-log[OH^-] = -log[\dfrac{0.01}{0.25}] = 1.four}\)

Then \(pH = {14 - pOH}\), plugging in pH = 14 - 1.four = 12.vi

Polyprotic Acids and Bases

So far, we have only considered monoprotic acids and bases, all the same there are various other substances that can donate or have more than than proton per molecule and these are known as polyprotic acids and bases. Polyprotic acids and bases have multiple dissociation constants, such equally \(K_{a1}\), \(K_{a2}\), \(K_{a3}\) or \(K_{b1}\), \(K_{b2}\), and \(K_{b3}\), and equivalence points depending on the number of times dissociation occurs.

Polyprotic Acids

Polyprotic acids are acids that can lose several protons per molecule. They can exist farther categorized into diprotic acids and triprotic acids, those which tin can donate ii and three protons, respectively. The best way to demonstrate polyprotic acids and bases is with a titration curve. A titration bend displays the multiple acid dissociation constants (\(K_a\)) as portrayed below.

Hither is a list of some common polyprotic acids:

Name	Formula	\(\pmb{K_{a1}}\)	\(\pmb{K_{a2}}\)	\(\pmb{K_{a3}}\)
Sulfuric acrid (stiff, diprotic)	H₂SO₄	1.0 10 10³	one.2 x ten^-two	-
Carbonic acid (weak, diprotic)	H₂CO_three	4.2 x ten^-7	four.8 10 10^-11	-
Phosphoric acid (weak, triprotic)	H_threePO₄	7.1 x 10^-3	half-dozen.3 10 10^-viii	4.two x x^-13

Polyprotic Bases

Polyprotic bases are bases that tin can attach several protons per molecule. Similar to polyprotic acids, polyprotic bases tin be categorized into diprotic bases and triprotic bases. Here is a list of some common polyprotic bases:

Name	Formula	\(\pmb{K_b}\)
Barium hydroxide (strong, diprotic)	Ba(OH)_two
Phosphate ion (triprotic)	PO_iv ³ ^-
Sulfate ion (diprotic)	SO_four ⁱⁱ ^-

Example \(\PageIndex{2}\)

For a 4.0 One thousand H_threePO_four solution, calculate (a) [H₃O⁺] (b) [HPO₄2-^-] and (c) [PO₄3^-].

\[H_3PO_4 + H_2O \Longleftrightarrow H_3O^+ + H_2PO_4^-\]

Solution

(a) Using Ice Tables y'all get:

\[K_{a1} = \dfrac{[H_3O^+][H_2PO_4^-]}{[H_3PO_4]}\]

So,

\(10^two\) = .0284

\(10\) = 0.17 M

(b) From part (a), \(10\) = [H₂PO₄ ^-] = [H₃O⁺] = 0.17 M

(c) To determine [H_iiiO⁺] and [H_twoPO₄ ^-], information technology was causeless that the 2d ionization constant was insignificant.

The new equation is as follows:

\(H_2PO_4^- + H_2O \Longleftrightarrow H_3O^+ + HPO_4^{2-}\)

Using ICE Tables once more:

\(K_{a2} = [HPO_4^{2-}] = 6.3 \times 10^{-8}\)

Example \(\PageIndex{3}\)

The polyprotic acid H₂And then_four can ionize 2 times ( \(K_{a1}>>one\), \(K_{a2} = 1.i * ten^-2\)). If we start with 9.50*10^-3 Yard solution of H₂Then_iv, what are the final concentrations of H₂So₄, HSO_four ^-, SO_iv ² ^-, and H_iiiO⁺.

Solution

The equation for the beginning ionization is \(H_2SO_4 + H_2O \Longleftrightarrow H_3O^+ + HSO_4^-\). This equation goes to completion because H_iiAnd then₄ is a stiff acrid and\(K_{a1}>>1\).

So since the reaction goes to completion, doing an Water ice Tabular array you lot get [H_iii0⁺] = 9.50*10^-three M and [HSO₄ ^-] = 9.l*10^-3 M (afterwards the starting time ionization).

The equation of the second ionization is \(HSO_4- + H_2O \Longleftrightarrow H_3O^+ + SO_4^ii-\). Using the equation \(K_{a2} = \dfrac{[H_3O^+][SO_4^2-^-]}{[HSO_4^-]}\), \(K_{a2} = 1.i * ten^-2\), and an ICE Table to get \(10^ii + .0.0205x - 0.0001045 = 0\).

Then you use the quadratic equation to solve for X, to get \(x\) = 0.004226.

Now we need to solve for the necessary concentrations

\([H_2S0_4]\) = 0 (considering the first ionization reaction went to completion)

\([HS0_4^-]\) = \(k_{a1}\) - \(k_{a2}\) = 9.l*ten^-3 M - 0.004226 Yard = v.27*x^-3 M

\([SO_4^2-]\) = \(k_{a2}\) = .004226 K

\([H_3O^+]\) = \(k_{a1}\) + \(k_{a2}\) = nine.l*10^-3 M + 0.004226 M = i.37*10^-ii M

Summary

K_a and Grand_b are equilibrium constants and a high value signifies a stronger acid or base.
Acrid are proton donors and bases are proton acceptors.
Monoprotic acrid/base corresponds to the donation/acceptance of only one proton.
Polyprotic acid/base of operations corresponds to the donation/credence of more ane proton.
- Call back diprotic and triprotic.
\(K_{a1}\)>\(K_{a2}\)>\(K_{a3}\)

Common Errors

Assuming that the [H₃0⁺] is the same for all the ionizations.
- In fact, the pH is dominated by merely the first ionization, but the later ionizations do contribute very slightly.

References

Petrucci, et al. General Chemical science: Principles & Modern Applications. 9th ed. Upper Saddle River, New Jersey: Pearson/Prentice Hall, 2007.
Sadava, et al. Life: The Science of Biology. 8th ed. New York, NY. W.H. Freeman and Company, 2007
Hulanicki, Adam. Reactions of Acids and Bases In Analytical Chemistry. New York, NY: Ellis Horowood Express, 1987.

Contributors and Attributions

Christopher Spohrer, Zach Wyatt (UCD)