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16.5: Acid-Base Equilibrium Calculations

  • Page ID
    60746
  • Introduction

    In this section we will apply equilibrium calculations from chapter 15 to Acids, Bases and their Salts. Although RICE diagrams can always be used, there are many conditions where the extent of ionization is so small that they can be simplified. After a review of the percent ionization, we will cover strong acids and bases, then weak acids and bases, and then acidic and basic salts. We will usually express the concentration of hydronium in terms of pH.

    Strong Acids

    For the generic reaction of a strong acid Ka is a large number meaning it is a product favored reaction that goes to completion and we use a one way arrow.
    \[HA(aq)+H_2O(l) \rightarrow H_3O^+(aq)+A^-(aq)\]

    The acid undergoes 100% ionization, meaning the equilibrium concentration of \([A^-]_{e}\) and \([H_3O^+]_{e}\) both equal the initial Acid Concentration \([HA]_{i}\), and so there is no need to use an equilibrium constant.

    Some common strong acids are HCl, HBr, HI, HNO3, HClO3 and HClO4.

    H2SO4 is often called a strong acid because the first proton is kicked off (Ka1=1x102), but the second is not 100% ionized (Ka2=1.0x10-2), but it is also not weak. We will cover sulfuric acid later when we do equilibrium calculations of polyatomic acids.

    NOTE: You do not need an Ionization Constant for these reactions

    Exercise \(\PageIndex{1}\)

    What is the pH of 0.025M HCl?

    Answer

    pH = -log \([H_3O^+]_{e}\) = -log0.025 = 1.60

    Strong Bases

    Just like strong acids, strong Bases 100% ionize (KB>>0) and you solve directly for pOH, and then calculate pH from pH + pOH =14.

    There are two basic types of strong bases, soluble hydroxides and anions that extract a proton from water. Some anions interact with more than one water molecule and so there are some polyprotic strong bases.

    Monoprotic Strong Bases

    Alkali metal hydroxides release hydroxide as their anion

    \[NaOH(aq)+H_2O(l) \rightarrow Na^+(aq)+OH^-(aq)\]

    Soluble hydrides release hydride ion to the water which reacts with the water forming hydrogen gas and hydroxide. This can be seen as a two step process.

    \[\begin{align}NaH(aq) & \rightarrow Na^+(aq)+H^-(aq) \nonumber \\ H^-(aq)+H_2O(l) &\rightarrow H_2(g)+OH^-(aq) \nonumber \\ \nonumber \\ \text{Net} & \text{ Equation} \nonumber \\ \nonumber \\ NaH(aq)+H_2O(l) & \rightarrow Na^+(aq) + H_2(g)+OH^-(aq) \end{align}\]

    Exercise \(\PageIndex{2}\)

    What is the pH of a solution made by dissolving 1.2g NaH into 2.0 liter of water?

    Answer

    1.2 g sodium hydride in two liters results in a 0.025M NaOH that would have a pOH of 1.6. From that the final pH is calculated using pH + pOH = 14.

    \[\frac{\left ( 1.2gNaH \right )}{2.0L}\left ( \frac{molNaH}{24.008g} \right )\left ( \frac{molOH^-}{molNaH} \right )=0.025M OH^- \\
    pOH=-log0.025=1.60 \\
    pH=14-pOH = 14-1.60 = 12.40 \nonumber \]
    Note this could have been done in one step
    \[pH=14+log(\frac{\left ( 1.2gNaH \right )}{2.0L}\left ( \frac{molNaH}{24.008g} \right )\left ( \frac{molOH^-}{molNaH} \right )) = 12.40 \nonumber\]

    Diprotic Strong Bases

    Soluble oxides react with water very vigorously to produce two hydroxides. For example CaO reacts with water to produce aqueous calcium hydroxide. This reaction has been used in chemical heaters and can release enough heat to cause water to boil.

    \[\begin{align}CaO(aq) &\rightarrow Ca^{+2}(aq)+O^{-2}(aq) \nonumber \\ O^{-2}(aq)+H_2O(l) &\rightarrow 2OH^-(aq) \nonumber \\ \nonumber \\ \text{Net} & \text{ Equation} \nonumber \\ \nonumber \\ CaO(aq)+H_2O(l) & \rightarrow Ca^{+2} + 2OH^-(aq) \end{align}\]

    Exercise \(\PageIndex{3}\)

    What is the pH of a solution made by dissolving 1.21g calcium oxide to a total volume of 2.00 L?

    Answer

    Add ans

    \[\dfrac{\left ( 1.21gCaO\right )}{2.00L}\left ( \frac{molCaO}{56.08g} \right )\left ( \frac{2molOH^-}{molCaO} \right )=0.0216M OH^- \\[5pt] pOH=-\log0.0216=1.666 \\[5pt] pH = 14-1.666 = 12.334 \nonumber \]

    Note this could have been done in one step

    \[pH=14+log(\frac{\left ( 1.21gCaO\right )}{2.00L}\left ( \frac{molCaO}{56.08g} \right )\left ( \frac{2molOH^-}{molCaO} \right)) = 12.334 \nonumber\]

    Triprotic Strong Bases

    Nitrides (N-3) react very vigorously with water to produce three hydroxides. For example Li3N reacts with water to produce aqueous lithium hydroxide and ammonia. This is a violent reaction, which makes sense as the [-3] charge is going to have a very strong pull on the hydrogens as it forms ammonia.

    \[\begin{align}Li_3N(aq) &\rightarrow 3Li^{+}(aq)+N^{-3}(aq) \nonumber \\ N^{-3}(aq)+3H_2O(l) &\rightarrow 3OH^-(aq) + NH_3(aq) \nonumber \\ \nonumber \\ \text{Net} & \text{ Equation} \nonumber \\ \nonumber \\ Li_3N(aq)+3H_2O(l) & \rightarrow 3Li^{+}(aq) + 3OH^-(aq)+ NH_3(aq) \end{align}\]

    Exercise \(\PageIndex{4}\)

    What is the pH of a solution made by dissolving 1.2g lithium nitride to a total volume of 2.0 L?

    Answer

    \[\frac{\left ( 1.2gLi_3N\right )}{2.0L}\left ( \frac{molLi_3N}{34.83g} \right )\left ( \frac{3molOH^-}{molLi_3N} \right )=0.0517M OH^- \\ pOH=-log0.0517=1.29 \\ pH = 14-1.29 = 12.71 \nonumber \]

    Note this could have been done in one step

    \[pH=14+log(\frac{\left ( 1.2gLi_3N\right )}{2.0L}\left ( \frac{molLi_3N}{34.83g} \right )\left ( \frac{3molOH^-}{molLi_3N} \right )) = 12.71 \nonumber\]

    Weak Acids

    There are two types of weak acid calculations, and these are analogous to the two type of equilibrium calculations we did in sections 15.3 and 15.4.

    • Type 1: Determine Ka for an unknown acid from initial concentration and final concentration of one species, typically hydrogen, as defined by the pH. This is analogous to the work in section 15.3, except that it is applied to acids.
    • Type2: Calculate final pH from initial concentration and Ka. This is analogous to section15.4 but is applied only to acids, and we will look for simplifications of the RICE diagram so we do not have to solve the quadratic equation.

    Calculating Ka

    In these problems you typically calculate the Ka of a solution of known molarity by measuring it's pH. You will want to be able to do this without a RICE diagram, but we will start with one for illustrative purpose.

    \[HA \rightleftharpoons H^+ + A^-\]

    Starting with an ICE diagram.

    \(HA\) \(H^+\) \(A^-\)
    Initial [HA]i 0 0
    Change -x +x +x
    Equilibrium [HA]i-x x x

    From the ice diagram it is clear that \[K_a =\frac{x^2}{[HA]_i-x}\] and you should be able to derive this equation for a weak acid without having to draw the RICE diagram. Noting that \(x=10^{-pH}\) and substituting, gives\[K_a =\frac{(10^{-pH})^2}{[HA]_i-10^{-pH}}\]

    Predicting pH

    The second type of problem is to predict the pH of a weak acid solution if you know Ka and the acid concentration. Ka values for many weak acids can be obtained from table 16.3.1 There are two cases

    • Case I: [HA]>100 Ka , then \(\large{H^{+}=\sqrt{K_{a}[HA]_{i}}}\) and \(\large{pH=-log\sqrt{K_{a}[HA]_{i}}}\)
      • In this case the percent ionized is small and so the amount ionized is negligible to the initial acid concentration. This shortcut used the complete the square technique and its derivation is below in the section on percent ionization, as it is only legitimate if the percent ionization is low.
    • Case2: HA\(\ngtr\)100Ka , in which case you need to solve the quadratic equation of the RICE diagram for x, noting x = [H+].
      • In this case the percent ionized is not negligible, and you can not use the approximation used in case 1. Review section 15.4 for case 2 problems.

    Note, the approximation [HA]>Ka is usually valid for two reasons, but realize it is not always valid.

    1. "Weak" in "weak acid" means Ka is small. That is the definition of a weak acid, (and a strong acid has a large Ka).
    2. Most acid concentrations in the real world are larger than Ka values, and there is a limit to how dilute an acid can get and still affect the pH. That is, as you dilute an acid the pH rises, but you can not raise it above 7, the value of neutral water. (Even strong acids have no effect when their concentration is below 10-7M.

    Percent Ionization

    The "Case 1" shortcut \([H^{+}]=\sqrt{K_{a}[HA]_{i}}\) avoided solving the quadratic formula for the equilibrium constant expression in the RICE diagram by removing the "-x" term from the denominator and allowing us to "complete the square". This is only valid if the percent ionization is so small that x is negligible to the initial acid concentration. We said this is acceptable if 100Ka <[HA]i. We will now look at this derivation, and the situations in which it is acceptable.

    \[ K_a =\underbrace{\frac{x^2}{[HA]_i-x}\approx \frac{x^2}{[HA]_i}}_{\text{true if x}<<[HA]_i} \]

    solving the simplified version for x and noting that [H+]=x, gives:
    \[\begin{align} x^2 & =K_a[HA]_i \nonumber \\ x & =\sqrt{K_a[HA]_i} \nonumber \\ [H^+] & =\sqrt{K_a[HA]_i}\end{align}\]

    To understand when the above shortcut is valid one needs to relate the percent ionization to the [HA]i >100Ka rule of thumb.

    Percent Ionization %I:

    \[\%I=\frac{ x}{[HA]_i}=\frac{ [A^-]}{[HA]_i}100\]

    One way to understand a "rule of thumb" is to apply it. Table 16.5.2 tabulates hydronium concentration for an acid with Ka=10-4 at three different concentrations, where [HA]i is greater than, less than or equal to 100 Ka. What is important to understand is that under the conditions for which an approximation is valid, and how that affects your results. In condition 1, where the approximation is valid, the short cut came up with the same answer for percent ionization (to three significant digits). In column 2 which was the limit, there was an error of .5% in percent ionization and the answer was valid to one sig. fig. but in case 3, which was clearly not valid, you got a completely different answer. To get a real feel for the problems with blindly applying shortcuts, try exercise 16.5.5, where [HA]i <<100Ka and the answer is complete nonsense.

    Column Relative magnitudes of [HA]i to Ka=10-4 [HA]i \(H^+=\frac{-b \pm\sqrt{b^{2}-4ac}}{2a}\) %Ionization
    (quadratic)
    \([H^+]=\sqrt{K_a[HA]_i}\) %Ionization
    (complete Square)
    1 [HA]i >100Ka 1 0.009950 0.995 0.01 1
    2 [HA]i =100Ka 10-2 9.5x10-4 9.5% 0.001 10
    3 [HA]i <100Ka 10-4 6.18x10-5 61.8% .0001 100%

    Table\(\PageIndex{2}\): Comparison of hydronium ion and percent ionizations for various concentrations of an acid with K Ka=10-4. In the above table, \(H^+=\frac{-b \pm\sqrt{b^{2}-4ac}}{2a}\) became \(H^+=\frac{-K_a \pm\sqrt{(K_a)^{2}+4K_a[HA]_i}}{2a}\). Note complete the square gave a nonsense answer for row three, as the criteria that [HA]i >100Ka was not valid.

    One other trend comes out of this table, and that is that the percent ionization goes up and concentration goes down.

    Exercise \(\PageIndex{5}\)

    When [HA]i >100Ka it is acceptable to use \([H^+] =\sqrt{K_a[HA]_i}\). Use this equation to calculate the percent ionization for a 1x10-6M solution of an acid with a Ka = 1x10-4M, and discuss (explain) the answer.

    Answer

    \[[H^+] =\sqrt{10^{-4}10^{-6}} = 10^{-5} \nonumber \]

    \[\%I=\frac{ x}{[HA]_i}=\frac{ [A^-]}{[HA]_i}100 \\ \frac{ 10^{-5}}{10^{-6}}100= 1,000 \% \nonumber \]

    The above answer is obvious nonsense and the reason is that the initial acid concentration greater than 100 times the ionization constant, in fact, it was less. So the percent ionization was not negligible and this problem had to be solved with the quadratic formula.

    Problems

    Exercise \(\PageIndex{6}\)

    What is the pH if 10.0 g Acetic Acid is diluted to 1.00 L?

    Answer

    From table 16.3.1 the value of K is determined to be 1.75x10-5 ,and acetic acid has a formula weight of 60.05g/mol, so

    \[[HC_2H_3O_2]=\left ( \frac{10.0gHC_2H_3O_2}{1.00L} \right )\left ( \frac{molHC_2H_3O_2f}{60.05g} \right )=0.167M \nonumber \]

    \[pH=-log\sqrt{1.75x10^{-5}[0.167]}=2.767.\]

    Weak Bases

    There are two types of weak base calculations, and these are analogous to the two type of equilibrium calculations we did in sections 15.3 and 15.4.

    • Type 1: Determine Kb for an unknown base from initial concentration and final concentration of one species, Typically the final concentration is expressed in terms of pH, but the equilibrium involves hydroxide not hydronium, and so you use the relationship pH+pOH=14, or [H3O+][OH-]=10-14 to relate hydroxide to hydronium.
    • Type2: Calculate final pH or pOH from initial concentrations and Kb. This is analogous to section15.4 but is applied only to bases, and we will look for simplifications of the RICE diagram so we do not have to solve the quadratic equation.

    Calculating Kb

    In these problems you typically calculate the Ka of a solution of know molarity by measuring it's pH. You will want to be able to do this without a RICE diagram, but we will start with one for illustrative purpose.

    \[B + H_2O \rightleftharpoons BH^+ + OH^-\]

    We will start with an ICE diagram, note, water is omitted from the equilibrium constant expression and ICE diagram because it is the solvent and thus its concentration is so much greater than the amount ionized, that it is essentially constant. This is similar to what we did in heterogeneous equilibiria where we omitted pure solids and liquids from equilibrium constants, but the logic is different (this is a homogeneous equilibria and water is the solvent, it is not a separate phase).

    \(B\) \(BH^+\) \(OH^-\)
    Initial [B]i 0 0
    Change -x +x +x
    Equilibrium [B]i-x x x

    From the ice diagram it is clear that

    \[K_b =\frac{x^2}{[B]_i-x}\]

    and you should be able to derive this equation for a weak acid without having to draw the RICE diagram.

    Noting that \(x=10^{-pOH}\) and substituting, gives

    \[K_b =\frac{(10^{-pOH})^2}{[B]_i-10^{-pOH}}\]

    Note, if you are given pH and not pOH, you simple convert to pOH, pOH=14-pH and substitute.

    Predicting pOH

    The second type of problem is to predict the pH or pOH for a weak base solution if you know Kb and the initial base concentration. Kb values for many weak bases can be obtained from table 16.3.2 There are two cases

    • Case I: [B]>100 Kb , then \(\large{OH^{-}=\sqrt{K_{b}[B]_{i}}}\) and \(\large{pOH=-log\sqrt{K_{b}[B]_{i}}}\)
      • In this case the percent ionized is small and so the amount ionized is negligible to the initial base concentration
    • Case2: B\(\ngtr\)100Kb , in which case you need to solve the quadratic equation of the RICE diagram for x, noting x = [OH-].
      • In this case the percent ionized is not negligible, and you can not use the approximation used in case 1. Review section 15.4 for case 2 problems.

    Note, the approximation [B]>Kb is usually valid for two reasons, but realize it is not always valid.

    1. "Weak" in "weak base" means Kb is small. That is the definition of a weak base, (and a strong base has a large Kb).
    2. Most base concentrations in the real world are larger than Kb values, and there is a limit to how dilute a base can get and still affect the pH (or pOH), as it can not reduce the pH below 7, the value of neutral water. (Even strong bases have no effect when their concentration is below 10-7M.)

    Exercise \(\PageIndex{7}\)

    What is the pH if 10.0 g Methyl Amine ( CH3NH2) is diluted to 1.00 L? (Obtain Kb from Table 16.3.1)

    Answer

    From Table 16.3.1 the value of Kb is determined to be 4.6x10-4 ,and methyl amine has a formula weight of 31.053 g/mol, so

    \[[CH_3NH_2]=\left ( \frac{10.0g[CH_3NH_2}{1.00L} \right )\left ( \frac{mol[CH_3NH_2}{31.053g} \right )=0.322M \nonumber \]

    \[pOH=-log\sqrt{4.6x10^{-4}[0.322]}=1.92 \\ pH=14-1.92=12.08.\]

    pH of Basic Salts

    Salts of a weak acid and a strong base form basic solutions because the conjugate base of the weak acid removes a proton from water. in section 16.4.2.3 we determined how to calculate the equilibrium constant for the conjugate base of a weak acid. The example of sodium fluoride was used, and it was noted that the sodium ion did not react with water, but the fluoride grabbed a proton and formed hydrofluoric acid. We used the relationship \(K_aK_b'=K_w\) for a acid/ conjugate base pair (where the prime designates the conjugate) to calculate the ionization constant for the anion. So there is a second step to these problems, in that you need to determine the ionization constant for the basic anion of the salt.

    so \[\large{K'_{b}=\frac{10^{-14}}{K_{a}}}\]

    If \( [A^-]_i >100K'_{b}\), then:

    \[[OH^-]=\sqrt{K'_b[A^-]_i}=\sqrt{\frac{K_w}{K_a}[A^-]_i} \\
    pOH=-log\left ( \sqrt{\frac{K_w}{K_a}[A^-]_i} \right ) \\
    pH=14-pOH \\
    pH = 14+log\left ( \sqrt{\frac{K_w}{K_a}[A^-]_i} \right )\]

    Exercise \(\PageIndex{8}\)

    What is the pH of a 0.100 M solution of sodium hypobromite? You can get Ka for hypobromous acid from Table 16.3.1 .

    Answer

    First calculate the hypobromite ionization constant, noting \(K_aK_b'=K_w\) and \(K^a = 2.8x10^{-9}\) for hypobromous acid

    \[\large{K_{b}^{'}=\frac{10^{-14}}{K_{a}} = \frac{10^{-14}}{2.8x10^{-9}}=3.6x10^{-6}}\]

    Since \([Br^-]<100K_a\)

    \[p[OH^-]=-log\sqrt{ (3.6x10^{-6})(0.100)} = 3.22 \\ pH=14-pOH = 14-3.22=11\]

    pH of Acidic Salts

    Salts of a weak base and a strong acid form acidic solutions because the conjugate acid of the weak base protonates water. In section 16.4.2.2 we determined how to calculate the equilibrium constant for the conjugate acid of a weak base. The example of ammonium chlorides was used, and it was noted that the chloride ion did not react with water, but the ammonium ion transferred a proton to water forming hydronium ion and ammonia. The equilibrium constant for the acidic cation was calculated from the relationship \(K'_aK_b=K_w\) for a base/ conjugate acid pair (where the prime designates the conjugate).

    The calculation of the pH for an acidic salt is similar to that of an acid, except that there is a second step, in that you need to determine the ionization constant for the acidic cation of the salt from the basic ionization constant of the base that could have formed it.

    so

    \[\large{K'_{a}=\frac{10^{-14}}{K_{b}}}\]

    If \( [BH^+]_i >100K'_{a}\), then:
    \[[H^+]=\sqrt{K'_a[BH^+]_i}=\sqrt{\frac{K_w}{K_b}[BH^+]_i} \\
    pH=-log\sqrt{\frac{K_w}{K_b}[BH^+]_i}\]

    Exercise \(\PageIndex{9}\)

    What is the pH of a 0.100 M solution of hydroxylammonium chloride (NH3OHCl), the chloride salt of hydroxylamine? You can get Kb for hydroxylamine from Table 16.3.2 .

    Answer

    First calculate the hydroxylammonium ionization constant, noting \(K'_aK_b=K_w\) and \(K_b = 8.7x10^{-9}\) for hydroxylamine.

    \[\large{K_{a}^{'}=\frac{10^{-14}}{K_{b}} = \frac{10^{-14}}{8.7x10^{-9}}=1.1x10^{-6}}\]

    Since \([NH_3OH^+]<100K_a\)

    \[p[H^+]=-log\sqrt{ (1.1x10^{-6})(0.100)} = 3.50 \]

    pH of Polyprotic Acids

    In section 15.1.2.2 we discussed polyprotic acids and bases, where there is an equilbiria existing between the acid, the acid salts and the salts. In a diprotic acid there are two species that can protonate water, the acid itself, and the ion formed when it loses one of the two acidic protons (the acid salt anion). Generically, this can be described by the following reaction equations:

    \[H_2A(aq) + H_2O)l) \rightleftharpoons HA^- (aq) + H_3O^+(aq) \text{, where } K_{a1}=\frac{[HA^-][H_3O^+]}{H_2A} \]

    \[ HA^-(aq) +H_2O(l) \rightleftharpoons A^{-2}(aq) + H_3O^+(aq) \text{, where } K_{a2}=\frac{[A^-2][H_3O^+]}{HA^-}\]

    Note, in the first equation we are removing a proton from a neutral molecule while in the second we are removing it from a negative anion. As the protons are being removed from what is essentially the same compound, coulombs law indicates that it is tougher to remove the second one because you are moving something positive away from a negative anion. This means the second ionization constant is always smaller than the first. Another way to look at that is through the back reaction. Would the proton be more attracted to HA- or A-2? As the attraction for the minus two is greater than the minus 1, the back reaction of the second step is greater, indicating a small K. So,

    \[\large{K_{a1}>K_{a2}}\]

    Note, not only can you determine the concentration of H+, but also OH-, H2A, HA- and A-2.

    Rule of Thumb: If \(\large{K_{a1}>1000K_{a2}}\) you can ignore the second ionization's contribution to the hydronium ion concentration, and if \([HA]_i>100K_{a1}\) the problem becomes fairly simple.

    Consider H2A:

    Only the first ionization contributes to the hydronium ion concentration as the second ionization is negligible.

    For first ionization

    \(H_2A(aq) +H_2O(l) \rightleftharpoons HA^-(aq) + H_3O^+(aq)\)
    \(H_2A\) \(HA^-\) \(H_3O^+\)
    Initial [H2A]i 0 0
    Change -x +x +x
    Equilibrium [H2A]i-x x x

    Since \(\large{K_{a1}>1000K_{a2}}\) the acid salt anion \(HA^-\) and \(H_3O^+\) concentrations come from the first ionization. That is, the amount of the acid salt anion consumed or the hydronium ion produced in the second step (below) is negligible and so the first step determines these concentrations.

    \[\large{[H^+]= [HA^-] = \sqrt{K_{a1}[H_2A]_i}}\]

    Knowing hydronium we can calculate hydorixde"
    \[[OH^-]=\frac{K_w}{[H^+]}\]

    Since the second ionization is small compared to the first, we can also calculate the remaining diprotic acid after the first ionization

    \[[H_2A]_e= [H_2A]_i-x\]

    For the second ionization we will use "y" for the extent of reaction, and "x" being the extent of reaction which is from the first ionization, and equal to the acid salt anion and the hydronium cation (from above)

    \(HA^-(aq) +H_2O(l) \rightleftharpoons A^{-2}(aq) + H_3O^+(aq)\)
    \(HA^-\) \(A^{-2}\) \(H_3O^+\)
    Initial x 0 x
    Change -y +y +y
    Equilibrium x-y y x+y

    \[\begin{align}K_{a2} & =\frac{[A^{-2}][H_3O^+]}{HA^-} \nonumber \\ & = \underbrace{\frac{[x+y][y]}{[x-y]} \approx \frac{[x][y]}{[x]}}_{\text{negliible second ionization (y<<x)}} \nonumber \\ K_{a2} &=y \nonumber \\ & \therefore \\ [A^{-2}] & = K_{a2} \nonumber \end{align}\]

    Exercise \(\PageIndex{10}\)

    Calculate the concentration of all species in 0.50 M carbonic acid. See Table 16.3.1 for Acid Ionization Constants.

    Answer

    From Table 16.3 Ka1 = 4.5x10-7 and Ka2 = 4.7x10-11 . So both [H2A]i 100>Ka1 and Ka1 >1000Ka2 .

    \[ [H^+] = [HA^-] = \sqrt {K_{a1}[H_2A]_i} \\ = \sqrt{(4.5x10^{-7})(0.50)} = 4.7x10^{-4}M \nonumber\]

    \[[OH^-]=\frac{10^{-14}}{4.74x10^{-4}}=2.1x10^{-11}M \nonumber\]

    \[[H_2A]_e= 0.5 - 0.00047 =0.50 \nonumber\]

    \[[A^{-2}]=K_{a2}=4.7x10^{-11}M \nonumber\]

    Contributors and Attributions

    • Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:

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