Evaluations of ph for Aqueous Solutions

Transcription

1 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. The ph value is defined as the negative logarithm (10 as the base) of : ph=-log 10 More accurately, should be the activity of protons, a H+. 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 K w =[OH - ]. It is unit-less constant at a given temperature. For example, at 25 o C, K w = Unless otherwise stated, the temperature is always assumed to be 25 o C. K w = [OH ] = (@ 25 ph = log 10 = 7(@ 25 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 = [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 (Na 3 PO 4 ), the MBE of P element is o C) o C) 1 M = c 0 (Na 3 PO 4 ) = [PO 4 3 ] + [HPO 4 2 ] + [H 2 PO 4 2 ] + [H 3 PO 4 ] where the c 0 (Na 3 PO 4 ) is called the analytical concentration of Na 3 PO 4. 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 Page 1 of 11

2 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 = [OH ] + [Cl ] = K w + c 0(HCl) [Analytical equation] If c 0 (HCl) > 10 6 M, the protons contributed from water can be ignored, thus the analytical equation is reduced to = [OH ] + [A ] [A ] = c 0 (HA) For poly-protic strong acid solutions, the relationship between and c 0 (H n A) is (ignoring water hydrolysis) = n[a ] = nc 0 (H n A) where n is proton number in the chemical formula of a poly-protic strong acid, H n A. For poly-protic strong base solutions, [OH ] needs to be determined first using similar equations and then convert it to using K w to calculate ph 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 = [OH ] + [A ] = K w + K [HA] a where K a is the acid constant of HA. Solving the above equation yields = K w + K a [HA] The [HA] is usually unknown and needs further elaborations: [HA] = [HA] c 0 (HA) c 0(HA) = c 0(HA)[HA] [HA] + [A ] = c 0(HA)[HA] [HA] + K = [H+ ]c 0 (HA) a[ha] K a + Substitute [HA] in = K w + K a [HA], = K w + K ac 0 (HA) K a + [Analytical Equation #1] Page 2 of 11

3 If K a c 0 (HA) > 20K w, the protons contributed from water can be ignored. The analytical equation is reduced to K a [HA] = K a (c 0 (HA) [A ]) = K a (c 0 (HA) [OH ]) K a (c 0 (HA) ) Note that the MBE equation is used to derive from the second term to the third term. Solving the above quadratic equation, = K a K a 2 + 4K a c 0 (HA) 5% approximation: if less than 5% of the acid molecules are dissociated, i.e., c 0 (HA) [A ] c 0 (HA), thus K a c 0 (HA) < , = K w + K a [HA] = K w + K a (c 0 (HA) [A ]) K w + K a c 0 (HA) K w + K a c 0 (HA) +#2 If ignoring water dissociation and applying the 5% approximation, the simplest equation is K a c 0 (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 using K w to obtain the ph 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, H 2 A, the PBE is = [OH ] + [HA ] + 2[A 2 ] = K w + K a1 [HA ] + 2K a1 K a2 [H 2 A] 2 Page 3 of 11

4 Solve the above quadratic equation yields Evaluations of ph for Aqueous Solutions Tianyu Liu = K a1 [H 2 A] (1 + 2K a2 ) Note that the right hand side of the analytical equation still contains. Ignoring the second dissociation step of H 2 A, i.e., 2K a2 weak di-protic acids), i.e., 1 + 2K a2 1, K a1 [H 2 A] 1 (which is usually true for most Analytical equations of the 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 using K w to obtain the ph. 2.2 Salts Strong acid-strong base salts Typical examples include NaCl, NaNO 3, KCl etc. Protons are all from water and thus the ph value should be 7 (@25 o C), independent of the analytical concentration 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 (NH 4 Cl), sodium sulfide (Na 2 S) 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 or 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 [(NH 4 ) 2 CO 3 ], ammonium acetate (NH 4 Ac) etc. If denoting the salt as AB (A cation and B anion), the PBE can be written as Page 4 of 11

5 [HB] + = [OH ] + [A ] Substitute [HB] = [H+ ][B ], K a (HB) [A ] = K a (HA) [HA] and [OH ] = K w, = K a(ha)[ha] + K w 1 + [B ]/K a (HB) 5% approximation: if less than 5% of the acid molecules are dissociated, i.e., K a(ha) < c 0 (AB) and K b(b) c 0 (AB) < , [HA] c 0 (AB) and [B ] c 0 (AB), thus the analytical equation can be reduced to = K a(ha)c 0 (AB) + K w 1 + c 0 (AB)/K a (HB) 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.] If K a c 0 (AB) > 20K w, the protons contributed from water can be ignored, thus the analytical equation is reduced to = K a(ha)c 0 (AB) 1 + c 0 (AB)/K a (HB) Approximation #3 If c 0 (AB)/K a (HB) 20, the denominator 1 + c 0 (AB)/K a (HB) c 0 (HA)/K a1, thus the analytical equation can be reduced to Page 5 of 11

6 = K a (HA)K a (HB) + K w c 0 (AB) [Approximation Equation #3] +#3 If both the approximations apply, the analytical equation can be reduced to its simplest form, i.e., = K a (HA)K a (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 c 0 (AB) 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 Plug in K a1, K a2 and K w + [H+ ][HA ] K a1 Rearranging the above quadratic equation, + [H 2 A] = [OH ] + [A 2 ] = K w + K [HA ] a2 = K a2[ha ] + K w 1 + [HA ]/K a1 If K a1 1000K a2, the dissociation and hydrolysis of HA - are both insignificant and can be ignored, thus [HA ] c 0 (NaHA), and the analytical equation can be reduced to its simplest form, = K a2c 0 (NaHA) + K w 1 + c 0 (NaHA)/K a1 Note that this approximation is very likely to be possible because usually the first de-protonation Page 6 of 11

7 step is much easier than the subsequent steps. The approximation #2 and #3 shown below are both built on #1. If K a c 0 (NaHA) > 20K w, the protons contributed from water can be ignored, thus the analytical equation is reduced to = K a2c 0 (NaHA) 1 + c 0 (NaHA)/K a1 Approximation #3 If c 0 (NaHA)/K a1 20, the denominator 1 + c 0 (NaHA)/K a1 c 0 (NaHA)/K a1, thus the analytical equation is reduced to = K a1 K a2 + K w c 0 (NaHA) [Approximation Equation #3] +#3 If both approximations apply, the analytical equation can be reduced to its simplest form, i.e., = K a1 K a2 [Approximation Equation #4] The above equation clearly illustrates that under proper approximations, the ph value of an acid salt solution does not depend on c 0 (NaHA). 2.3 Mixtures 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 Substitute [A ] = K ac 0 (HA) K a + = c 0 (HCl) + [A ] + [OH ] (see Section for the derivation) and [OH ] = K w, Page 7 of 11

8 = c 0 (HCl) + K ac 0 (HA) K a + + K w Note that in order to solve the ph, a quadratic equation about needs to be solved because appears on both sides of the analytical equation. If c 0 (HCl) > 10 6 M or K a c 0 (HA) > 20K w, the protons contributed from water can be ignored, thus the analytical equation is reduced to = c 0 (HCl) + K ac 0 (HA) K a + Note that in order to solve the ph, a quadratic equation about needs to be solved because appears on both sides of the approximation equation #1. If c 0 (HCl) > K ac 0 (HA), the protons contributed from HA can be ignored, so as to the protons K a + from water (since the concentration of the strong acid is high), the analytical equation can be reduced to its simplest form, i.e., c 0 (HCl) 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 Substitute [A ] = K a(ha)[ha] = [A ] + [B ] + [OH ] = K a(ha)c 0 (HA), K a (HA)+ [B ] = K a(hb)[hb] = K a(hb)c 0 (HB), and K a (HB)+ [OH ] = K w = K a(ha)c 0 (HA) K a (HA) + + K a(hb)c 0 (HB) K a (HB) + + K w Note that in order to solve the ph, a quadratic equation about needs to be solved because appears on both sides of the analytical equation. Page 8 of 11

9 5% approximation: if less than 5% of the acid molecules are dissociated, i.e., K a(ha) < c 0 (HA) and K a(hb) < 2. 5 c 0 (HB) 10 3, [HA] c 0 (HA) and [HB] c 0 (HB), thus [A ] = K a(ha)[ha] K a (HA)c 0 (HA) and [B ] = K a(hb)[hb] K a(hb)c 0 (HB), an approximation equation can be derived = K a (HA)c 0 (HA) + K a (HB)c 0 (HB) + K w If K a (HA)c 0 (HA) > 20K w, and K a (HB)c 0 (HB) > 20K w, the protons contributed from water hydrolysis can be ignored, thus the analytical equation is reduced to = K a(ha)c 0 (HA) K a (HA) + + K a(hb)c 0 (HB) K a (HB) + Note that in order to solve the ph, a quadratic equation about needs to be solved because appears on both sides of the approximation equation #2. +#2 If the above approximations both apply, the analytical equation can be reduced to = K a (HA)c 0 (HA) + K a (HB)c 0 (HB) [Approximation Equation #3] Specifically, if one acid is much stronger than the other one, i.e., K a (HA)c 0 (HA) 20K a (HB)c 0 (HB), the analytical equation can be further reduced to its simplest form, i.e., = K a (HA)c 0 (HA) [Approximation Equation #4] 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 Page 9 of 11

10 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 The charge balance equation (CBE) is: [HA] + [A ] = c 0 (HA) + c 0 (A ) + [Na + ] = [OH ] + [A ] Note that [Na + ] = c 0 (A ). Solving the MBE and CBE altogether and obtain The can be expressed as [HA] = c 0 (HA) + [OH ] [A ] = c 0 (A ) + [OH ] = [HA] [A ] K a = c 0(HA) + [OH ] c 0 (A ) + [OH ] K a Note that in order to solve the ph, a quadratic equation about needs to be solved because appears on both sides of the analytical equation. If the acid, HA is not too diluted, i.e., c 0 (HA) [OH ], the analytical equation can be reduced to [H + c 0 (HA) ] c 0 (A )c 0 (A ) + [OH ] K a Note that in order to solve the ph, a quadratic equation about needs to be solved because appears on both sides of the approximation equation #1. If the conjugated base, NaA is not too diluted, i.e., c 0 (A ) [OH ], the analytical equation can be reduced to c 0(HA) + [OH ] c 0 (A K ) a Note that in order to solve the ph, a quadratic equation about needs to be solved because appears on both sides of the approximation equation #2. Page 10 of 11

11 +#2 If both the acid and the conjugated base are not too diluted, i.e., c 0 (HA) [OH ] and c 0 (A ) [OH ], the analytical equation can be reduced to its simplest form, i.e., c 0(HA) c 0 (A ) K a [Approximation Equation #3] Note that in order to solve the ph, a quadratic equation about needs to be solved because appears on both sides of the approximation equation #3. ==THE END== Page 11 of 11