Download Evaluations of pH for Aqueous Solutions

Evaluations of pH for Aqueous Solutions | Tianyu Liu | 2016.11@UCSC
Evaluations of pH for Aqueous Solutions
1. Preface
The key to calculate pH value of a certain solution is to determine the concentration of
hydrogen ions (or protons) at equilibrium, denoted as [H+]. The pH value is defined as the negative
logarithm (10 as the base) of [H+]:
pH=-log10 [H+]
More accurately, [H+] should be the activity of protons, aH+. To simplify our discussion, we will
assume the activity coefficients of all species are one, so activities equal concentrations.
Three important equations will be frequently used in the following discussion sections, i.e., ion
product of water, proton balance equation and mass balance equation.
Ion product of water is defined as Kw=[H+][OH-]. It is unit-less constant at a given temperature.
For example, at 25 oC, Kw=10-14. Unless otherwise stated, the temperature is always assumed to be
25 oC.
𝐾𝑤 = [OH − ] ∙ [H + ] = 10−14 (@ 25 oC)
pH = −log10 [H + ] = 7(@ 25 oC)
Proton balance equation, or PBE, is an equation states that the concentration of protons plus
concentration of any other specie(s) that accept(s) protons should always be equal to the
concentration of any other specie(s) that lose(s) protons. For example, in a solution containing
acetic acid (HAc), the PBE is
[H + ] = [OH − ] + [Ac − ]
Mass balance equation, or MBE, is an equation based on the mass conservation law. For
example, in a solution containing 1 M sodium phosphate (Na3PO4), the MBE of P element is
2−
2−
1 M = 𝑐0 (Na3 PO4 ) = [PO3−
4 ] + [HPO4 ] + [H2 PO4 ] + [H3 PO4 ]
where the 𝑐0 (Na3 PO4 ) is called the analytical concentration of Na3 PO4 . Species in “[]” represent
corresponding concentrations at equilibrium.
Besides the aforementioned three important equations, various approximations are usually
applied to simplify equations.
2. Calculations
2.1 Acids and Bases
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Evaluations of pH for Aqueous Solutions | Tianyu Liu | 2016.11@UCSC
2.1.1 Strong acids or bases
Protons in a strong acid solution are from both the acid itself and water. For a strong monoacid
solution, e.g., HCl, the PBE is
𝐾
[H + ] = [OH − ] + [Cl− ] = [H𝑤+] + 𝑐0 (HCl)
[Analytical equation]
Approximation #1
If 𝒄𝟎 (𝐇𝐂𝐥) > 𝟏𝟎−𝟔 𝐌, the protons contributed from water can be ignored, thus the analytical
equation is reduced to
[H + ] = [OH − ] + [A− ] ≈ [A− ] = 𝑐0 (HA)
[Approximation Equation #1]
For poly-protic strong acid solutions, the relationship between [H + ] and 𝑐0 (H𝑛 A) is
(ignoring water hydrolysis)
[H + ] = 𝑛[A− ] = 𝑛𝑐0 (H𝑛 A)
where n is proton number in the chemical formula of a poly-protic strong acid, H𝑛 A. For poly-protic
strong base solutions, [OH − ] needs to be determined first using similar equations and then convert
it to [H + ] using 𝐾𝑤 to calculate pH.
2.1.2 Weak mono-protic acids or mono-protic bases
Protons in a weak mono-protic acid solution are from both the acid itself and water. For a weak
mono-protic acid solution, HA, the PBE is
[HA]
𝐾𝑤
+ 𝐾𝑎 +
+
[H ]
[H ]
where Ka is the acid constant of HA. Solving the above equation yields
[H + ] = [OH − ] + [A− ] =
[H + ] = √𝐾𝑤 + 𝐾𝑎 [HA]
The [HA] is usually unknown and needs further elaborations:
[HA]
𝑐0 (HA)[HA]
𝑐0 (HA)[HA]
[H + ]𝑐0 (HA)
[HA] =
× 𝑐0 (HA) =
=
=
𝐾 [HA]
[HA] + [A− ]
𝑐0 (HA)
𝐾𝑎 + [H + ]
[HA] + 𝑎 +
[H ]
+
Substitute [HA] in [H ] = √𝐾𝑤 + 𝐾𝑎 [HA],
[H + ] = √𝐾𝑤 +
𝐾𝑎 [H + ]𝑐0 (HA)
𝐾𝑎 + [H + ]
[Analytical Equation #1]
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Approximation #1
If 𝑲𝒂 𝒄𝟎 (𝐇𝐀) > 𝟐𝟎𝑲𝒘 , the protons contributed from water can be ignored. The analytical equation
is reduced to
[H + ] ≈ √𝐾𝑎 [HA] = √𝐾𝑎 (𝑐0 (HA) − [A− ]) = √𝐾𝑎 (𝑐0 (HA) − [𝐻 + ] − [OH − ])
≈ √𝐾𝑎 (𝑐0 (HA) − [𝐻 + ])
Note that the MBE equation is used to derive from the second term to the third term. Solving the
above quadratic equation,
𝐾𝑎 1
+ √𝐾𝑎2 + 4𝐾𝑎 𝑐0 (HA)
2 2
[Approximation Equation #1]
[H + ] = −
Approximation #2
5% approximation: if less than 5% of the acid molecules are dissociated, i.e.,
𝑲𝒂
𝒄𝟎 (𝐇𝐀)
< 𝟐. 𝟓 × 𝟏𝟎−𝟑,
𝑐0 (HA) − [A− ] ≈ 𝑐0 (HA), thus
[H + ] = √𝐾𝑤 + 𝐾𝑎 [HA] = √𝐾𝑤 + 𝐾𝑎 (𝑐0 (HA) − [A− ]) ≈ √𝐾𝑤 + 𝐾𝑎 𝑐0 (HA)
[H + ] ≈ √𝐾𝑤 + 𝐾𝑎 𝑐0 (HA)
[Approximation Equation #2]
Approximation #1+#2
If ignoring water dissociation and applying the 5% approximation, the simplest equation is
[H + ] ≈ √𝐾𝑎 𝑐0 (HA)
[Approximation Equation #3]
For weak mono-protic base solutions, [OH − ] needs to be first determined using the similar
equations and then convert it to [H + ] using 𝐾𝑤 to obtain the pH.
2.1.3 Weak di-protic acids or di-protic bases
The protons in a weak di-protic acid solution are from both the acid itself and water. For a weak
di-protic solution, H2A, the PBE is
[H + ] = [OH − ] + [HA− ] + 2[A2− ] =
[HA− ] 2𝐾𝑎1 ∙ 𝐾𝑎2 [H2 A]
𝐾𝑤
+
𝐾
+
𝑎1
[H + ]
[H + ]
[H + ]2
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Solve the above quadratic equation yields
[H + ] = √𝐾𝑎1 [H2 A] (1 +
2𝐾𝑎2
)
[H + ]
[Analytical Equation]
Note that the right hand side of the analytical equation still contains [H + ].
Approximation #1
Ignoring the second dissociation step of H2A, i.e.,
𝟐𝑲𝒂𝟐
[𝐇 + ]
≪ 𝟏 (which is usually true for most
2𝐾
weak di-protic acids), i.e., 1 + [H+𝑎2] ≈ 1,
[H + ] ≈ √𝐾𝑎1 [H2 A]
[Approximation Equation #1]
Analytical equations of the [H + ] of poly-protic acids involve significantly more terms than
those of di-protic acids. But since the approximation #1 is usually valid, they all can be reduced to
the Approximation Equation #1 shown above.
Similarly for weak poly-protic bases, [OH − ] needs to be determined first using the similar
equations and then convert it to [H + ] using 𝐾𝑤 to obtain the pH.
2.2 Salts
2.2.1 Strong acid-strong base salts
Typical examples include NaCl, NaNO3, KCl etc. Protons are all from water and thus the pH
value should be 7 (@25 oC), independent of the analytical concentration.
2.2.2 Strong acid-weak base salts or strong base-weak acid salts
This type of salt is formed by mixing either a strong acid with a weak base or a strong base
with a weak acid. Typical examples include sodium acetate (NaAc), ammonium chloride (NH4Cl),
sodium sulfide (Na2S) etc. Since either the anions or the cations are weak bases or weak acids,
respectively, their pH values can be evaluated according to the equations derived in Section 2.1.2 or
2.1.3.
2.2.3 Weak acid-weak base salts
This type of salt is formed by mixing a weak acid with a weak base. Typical examples include
ammonium carbonate [(NH4)2CO3], ammonium acetate (NH4Ac) etc. If denoting the salt as AB (A –
cation and B –anion), the PBE can be written as
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[HB] + [H + ] = [OH − ] + [A− ]
Substitute [HB] =
[H+ ][B− ]
𝐾𝑎
[HA]
𝐾
, [A− ] = 𝐾𝑎 (HA) [H+] and [OH − ] = [H𝑤+],
(HB)
𝐾𝑎 (HA)[HA] + 𝐾𝑤
[H + ] = √
1 + [B − ]/𝐾𝑎 (HB)
[Analytical Equation]
Approximation #1
5% approximation: if less than 5% of the acid molecules are dissociated, i.e.,
𝑲𝒃 (𝐁)
𝟎 (𝐀𝐁)
and 𝒄
𝑲𝒂 (𝐇𝐀)
𝒄𝟎 (𝐀𝐁)
< 𝟐. 𝟓 × 𝟏𝟎−𝟑
< 𝟐. 𝟓 × 𝟏𝟎−𝟑, [HA] ≈ 𝑐0 (AB) and [B − ] ≈ 𝑐0 (AB), thus the analytical equation can be
reduced to
𝐾𝑎 (HA)𝑐0 (AB) + 𝐾𝑤
[H + ] = √
1 + 𝑐0 (AB)/𝐾𝑎 (HB)
[Approximation Equation #1]
Note that this approximation is always valid when the salt solution is not extremely diluted. The
approximation #2 and #3 shown below are both built on #1. [If the salt concentration is very small,
the analytical equation becomes messy and it is in general not recommended to use the equation for
pH evaluations. Use a piece of pH paper or a pH meter.]
Approximation #2
If 𝑲𝒂 𝒄𝟎 (𝐀𝐁) > 𝟐𝟎𝑲𝒘 , the protons contributed from water can be ignored, thus the analytical
equation is reduced to
𝐾𝑎 (HA)𝑐0 (AB)
[H + ] = √
1 + 𝑐0 (AB)/𝐾𝑎 (HB)
[Approximation Equation #2]
Approximation #3
If 𝒄𝟎 (𝐀𝐁)/𝑲𝒂 (𝐇𝐁) ≥ 𝟐𝟎 , the denominator 1 + 𝑐0 (AB)/𝐾𝑎 (HB) ≈ 𝑐0 (HA)/𝐾𝑎1 , thus the
analytical equation can be reduced to
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[H + ] = √𝐾𝑎 (HA)𝐾𝑎 (HB) +
𝐾𝑤
𝑐0 (AB)
[Approximation Equation #3]
Approximation #2+#3
If both the approximations apply, the analytical equation can be reduced to its simplest form,
i.e.,
[H + ] = √𝐾𝑎 (HA)𝐾𝑎 (HB)
[Approximation Equation #4]
The above equation clearly illustrates that under proper approximations, the pH value of a weak
acid-weak base salt does not depend on 𝑐0 (AB).
2.2.4 Acid salts
Protons in an acid salt solution are from both the salt itself and water. For an acid salt solution,
NaHA, the PBE is
[H + ] + [H2 A] = [OH − ] + [A2− ]
Plug in 𝐾𝑎1 , 𝐾𝑎2 and 𝐾𝑤
[H + ][HA− ]
[HA− ]
𝐾𝑤
= + + 𝐾𝑎2
[H ]
[H + ]
𝐾𝑎1
Rearranging the above quadratic equation,
[H + ] +
[H + ] = √
𝐾𝑎2 [HA− ] + 𝐾𝑤
1 + [HA− ]/𝐾𝑎1
[Analytical Equation]
Approximation #1
If 𝑲𝒂𝟏 ≫ 𝟏𝟎𝟎𝟎𝑲𝒂𝟐 , the dissociation and hydrolysis of HA- are both insignificant and can be
ignored, thus [HA− ] ≈ 𝑐0 (NaHA), and the analytical equation can be reduced to its simplest form,
𝐾𝑎2 𝑐0 (NaHA) + 𝐾𝑤
[H + ] = √
1 + 𝑐0 (NaHA)/𝐾𝑎1
[Approximation Equation #1]
Note that this approximation is very likely to be possible because usually the first de-protonation
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step is much easier than the subsequent steps. The approximation #2 and #3 shown below are both
built on #1.
Approximation #2
If 𝑲𝒂 𝒄𝟎 (𝐍𝐚𝐇𝐀) > 𝟐𝟎𝑲𝒘 , the protons contributed from water can be ignored, thus the analytical
equation is reduced to
[H + ] = √
𝐾𝑎2 𝑐0 (NaHA)
1 + 𝑐0 (NaHA)/𝐾𝑎1
[Approximation Equation #2]
Approximation #3
If 𝒄𝟎 (𝐍𝐚𝐇𝐀)/𝑲𝒂𝟏 ≥ 𝟐𝟎 , the denominator 1 + 𝑐0 (NaHA)/𝐾𝑎1 ≈ 𝑐0 (NaHA)/𝐾𝑎1 , thus the
analytical equation is reduced to
[H + ] = √𝐾𝑎1 𝐾𝑎2 +
𝐾𝑤
𝑐0 (NaHA)
[Approximation Equation #3]
Approximation #2+#3
If both approximations apply, the analytical equation can be reduced to its simplest form, i.e.,
[H + ] = √𝐾𝑎1 𝐾𝑎2
[Approximation Equation #4]
The above equation clearly illustrates that under proper approximations, the pH value of an acid
salt solution does not depend on 𝑐0 (NaHA).
2.3 Mixtures
2.3.1 Mixture of a strong acid (HCl) and a weak acid (HA)
Protons in a mixture of strong and weak acids come from HCl, HA and water. The PBE can be
written as
[H + ] = 𝑐0 (HCl) + [A− ] + [OH − ]
Substitute [A− ] =
𝐾𝑎 𝑐0 (HA)
𝐾𝑎
+[H+ ]
𝐾
(see Section 2.1.2 for the derivation) and [OH − ] = [H𝑤+],
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𝐾𝑎 𝑐0 (HA)
𝐾𝑤
+ +
+
𝐾𝑎 + [H ] [H ]
[Analytical Equation]
[H + ] = 𝑐0 (HCl) +
Note that in order to solve the pH, a quadratic equation about [H+] needs to be solved because
[H+] appears on both sides of the analytical equation.
Approximation #1
If 𝒄𝟎 (𝐇𝐂𝐥) > 𝟏𝟎−𝟔 𝐌 or 𝑲𝒂 𝒄𝟎 (𝐇𝐀) > 𝟐𝟎𝑲𝒘 , the protons contributed from water can be ignored,
thus the analytical equation is reduced to
𝐾𝑎 𝑐0 (HA)
𝐾𝑎 + [H + ]
[Approximation Equation #1]
[H + ] = 𝑐0 (HCl) +
Note that in order to solve the pH, a quadratic equation about [H+] needs to be solved because
[H+] appears on both sides of the approximation equation #1.
Approximation #2
If 𝒄𝟎 (𝐇𝐂𝐥) >
𝑲𝒂 𝒄𝟎 (𝐇𝐀)
𝑲𝒂 +[𝐇+ ]
, the protons contributed from HA can be ignored, so as to the protons
from water (since the concentration of the strong acid is high), the analytical equation can be
reduced to its simplest form, i.e.,
[H + ] ≈ 𝑐0 (HCl)
[Approximation Equation #2]
2.3.2 Mixture of two weak acids (HA+HB)
Protons in a mixture of two weak acids are from HA, HB and water. The PBE can be written as
[H + ] = [A− ] + [B− ] + [OH − ]
Substitute [A− ] =
𝐾𝑎 (HA)[HA]
[H+ ]
=
𝐾𝑎 (HA)𝑐0 (HA)
[H + ] =
𝐾𝑎
(HA)+[H+ ]
, [B− ] =
𝐾𝑎 (HB)[HB]
[H+ ]
=
𝐾𝑎 (HB)𝑐0 (HB)
𝐾𝑎
(HB)+[H+ ]
𝐾
, and [OH − ] = [H𝑤+]
𝐾𝑎 (HA)𝑐0 (HA) 𝐾𝑎 (HB)𝑐0(HB)
𝐾𝑤
+
+
𝐾𝑎 (HA) + [H + ] 𝐾𝑎 (HB) + [H + ] [H + ]
[Analytical Equation]
Note that in order to solve the pH, a quadratic equation about [H+] needs to be solved because
[H+] appears on both sides of the analytical equation.
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Approximation #1
5% approximation: if less than 5% of the acid molecules are dissociated, i.e.,
and
𝑲𝒂 (𝐇𝐁)
𝒄𝟎 (𝐇𝐁)
< 𝟐. 𝟓 × 𝟏𝟎−𝟑 ,
𝐾𝑎 (HA)𝑐0 (HA)
[H+ ]
and [B− ] =
[HA] ≈ 𝑐0 (HA) and [HB] ≈ 𝑐0 (HB) ,
𝐾𝑎 (HB)[HB]
[H+ ]
≈
thus
𝑲𝒂 (𝐇𝐀)
𝒄𝟎 (𝐇𝐀)
< 𝟐. 𝟓 × 𝟏𝟎−𝟑
[A− ] =
𝐾𝑎 (HA)[HA]
[H+ ]
≈
𝐾𝑎 (HB)𝑐0 (HB)
[H+ ]
, an approximation equation can be derived
[H + ] = √𝐾𝑎 (HA)𝑐0 (HA) + 𝐾𝑎 (HB)𝑐0 (HB) + 𝐾𝑤
[Approximation Equation #1]
Approximation #2
If 𝑲𝒂 (𝐇𝐀)𝒄𝟎 (𝐇𝐀) > 𝟐𝟎𝑲𝒘 , and 𝑲𝒂 (𝐇𝐁)𝒄𝟎 (𝐇𝐁) > 𝟐𝟎𝑲𝒘 , the protons contributed from
water hydrolysis can be ignored, thus the analytical equation is reduced to
𝐾𝑎 (HA)𝑐0 (HA) 𝐾𝑎 (HB)𝑐0 (HB)
+
𝐾𝑎 (HA) + [H + ] 𝐾𝑎 (HB) + [H + ]
[Approximation Equation #2]
[H + ] =
Note that in order to solve the pH, a quadratic equation about [H+] needs to be solved because
[H+] appears on both sides of the approximation equation #2.
Approximation #1+#2
If the above approximations both apply, the analytical equation can be reduced to
[H + ] = √𝐾𝑎 (HA)𝑐0 (HA) + 𝐾𝑎 (HB)𝑐0 (HB)
[Approximation Equation #3]
Specifically,
if
one
acid
is
much
stronger
than
the
other
one,
i.e., 𝑲𝒂 (𝐇𝐀)𝒄𝟎 (𝐇𝐀) ≥ 𝟐𝟎𝑲𝒂 (𝐇𝐁)𝒄𝟎 (𝐇𝐁), the analytical equation can be further reduced to its
simplest form, i.e.,
[H + ] = √𝐾𝑎 (HA)𝑐0 (HA)
[Approximation Equation #4]
2.3.3 Mixtures of conjugated acids and bases
Here we will only discuss the weak mono-protic acids mixed with their conjugated bases. For
poly-protic strong acids, their mixtures’ pH values are always 7. For poly-protic weak acids, only
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their first de-protonation steps are considered, so they are treated as weak mono-protic acids. All
equations that are needed for pH values of poly-protic weak acids are identical as the ones derived
below.
Consider a mixed system containing HA and NaA, the MBE can be written as
[HA] + [A− ] = 𝑐0 (HA) + 𝑐0 (A− )
The charge balance equation (CBE) is:
[H + ] + [Na+ ] = [OH − ] + [A− ]
Note that [Na+ ] = 𝑐0 (A− ). Solving the MBE and CBE altogether and obtain
[HA] = 𝑐0 (HA) − [H + ] + [OH − ]
[A− ] = 𝑐0 (A− ) + [H + ] − [OH − ]
The [H + ] can be expressed as
[H + ] =
[HA]
𝑐0 (HA) − [H + ] + [OH − ]
𝐾
=
𝐾
[A− ] 𝑎 𝑐0 (A− ) + [H + ] − [OH − ] 𝑎
[Analytical Equation]
Note that in order to solve the pH, a quadratic equation about [H+] needs to be solved because
[H+] appears on both sides of the analytical equation.
Approximation #1
If the acid, HA is not too diluted, i.e., 𝒄𝟎 (𝐇𝐀) ≫ [𝐎𝐇 − ] − [𝐇 + ], the analytical equation can be
reduced to
[H + ] ≈
𝑐0 (HA)
−
−
𝑐0 (A )𝑐0 (A ) + [H + ]
− [OH − ]
[Approximation Equation #1]
𝐾𝑎
Note that in order to solve the pH, a quadratic equation about [H+] needs to be solved because
[H+] appears on both sides of the approximation equation #1.
Approximation #2
If the conjugated base, NaA is not too diluted, i.e., 𝒄𝟎 (𝐀− ) ≫ [𝐇 + ] − [𝐎𝐇 − ], the analytical
equation can be reduced to
𝑐0 (HA) − [H + ] + [OH − ]
𝐾𝑎
𝑐0 (A− )
[Approximation Equation #2]
[H + ] ≈
Note that in order to solve the pH, a quadratic equation about [H+] needs to be solved because
[H+] appears on both sides of the approximation equation #2.
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Approximation #1+#2
If both the acid and the conjugated base are not too diluted, i.e., 𝒄𝟎 (𝐇𝐀) ≫ [𝐎𝐇 − ] − [𝐇 + ]
and 𝒄𝟎 (𝐀− ) ≫ [𝐇 + ] − [𝐎𝐇 − ], the analytical equation can be reduced to its simplest form, i.e.,
𝑐0 (HA)
𝐾
𝑐0 (A− ) 𝑎
[Approximation Equation #3]
[H + ] ≈
Note that in order to solve the pH, a quadratic equation about [H+] needs to be solved because
[H+] appears on both sides of the approximation equation #3.
==THE END==
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