# Chapter 2

## SOLVING ACID/BASE AND BUFFER PROBLEMS

All acid-base problems involving aqueous solutions of weak acids (HA) and/or their corresponding base forms ($$\mathrm{A^{-}}$$) fall into three distinct classes.

Class 1: Solutions of "Acid Form" only, e.g., the pH of 0.01 M HA = ?

Class 2: Solutions of "Base Form" only, e.g., the pH of 0.01 M $$\mathrm{A^{-}}$$ = ?

Class 3: Solutions of both "Acid" and "Base" forms, e.g., the pH of a solution containing 0.01 M HA and 0.01 M $$\mathrm{A^{-}}$$ = ?

Methods for solving each class of problem are described below. To solve problems involving buffers and titration, simply determine which class a problem falls into and solve accordingly.

For all three classes of problems, two equilibrium conditions must be satisfied:

$\mathrm{EQ1: \: HA \Leftrightarrow H^{+}+A^{-}; \: K_{a}=\dfrac{[H^{+}][A^{-}]}{[HA]}}$

$\mathrm{EQ2: \: H_{2}O \Leftrightarrow H^{+}+OH^{-}; \: K_{eq}=\dfrac{[H^{+}][A^{-}]}{[HA]}}$

For the dissociation of pure water

$\mathrm{EQ3: \: K_{eq}=\dfrac{[H^{+}][OH^{-}]}{[H_{2}O]}}=1.8\times 10^{-16}$

Since the concentration of pure water is large, $$\mathrm{[H_{2}O] = 55.6\: M}$$, its concentration remains unchanged during its dissociation. Hence a new dissociation constant $$\mathrm{K_{W}}$$ is defined.

$\mathrm{EQ4: \: K_{W}=K_{eq}[H_{2}O]=[H^{+}][OH^{-}]}=10^{-14}$

This equality is valid for most dilute solutions of weak acids and/or bases and is important for solving class 1-3 acid base problems.

### Class 1 Adding only the acid (HA) form.

What is the pH of a 0.01 M solution of a weak acid HA with a $$\mathrm{pk_{a}}$$ of 5.0?

Solution:

This is a class 1 problem since initially only the "acid form" of HA is present.

$$\mathrm{H}^{+}$$ ions arise from the dissociation of HA and from the dissociation of $$\mathrm{H_{2}O}$$. Water is a very weak acid, and most of the $$\mathrm{H^{+}}$$ will come from the HA dissociation (except at very low HA concentration), and thus the $$\mathrm{H_{2}O}$$ dissociation can be ignored. Each X moles of HA that dissociates give X mole of $$\mathrm{H^{+}}$$ and X mole of $$\mathrm{A^{-}}$$. HA will dissociate until equilibrium is achieved.

$\mathrm{HA \Leftrightarrow H^{+}+A^{-}}\: \mathrm{pK_{a}}=5.0; \: \mathrm{K_{a}}=10^{-5}$

HA $$\mathrm{H^{+}}$$ $$\mathrm{A^{-}}$$
Initial conditions $$0.01\: \mathrm{M}$$ $$\sim 0$$ $$\sim 0$$
Change due to dissociation $$-x$$ $$+x$$ $$+x$$
Net concentration at equilibrium $$0.01-x$$ $$x$$ $$x$$

This leads to the following equation:

$\mathrm{K_{a}=\dfrac{[H^{+}][A^{-}]}{[HA]}}=\dfrac{x\cdot x}{0.01-x}=\dfrac{x^{2}}{0.01-x}=10^{-5}$

In general, the equation shown above is a quadratic equation and can be solved exactly for $$x$$. If, however, $$x$$ is much less than [HA] initial, it can be neglected in the denominator and the equation is simplified. For most situations, $$x$$ can be ignored whenever it is less than 5% - 10% of HA at equilibrium. In a solution of a weak acid alone, this occurs when $$\mathrm{[HA]_{initial}>100\cdot K_{a}}$$.

For this problem, $$\mathrm{[HA]_{initial} = 0.01\: M = 1000\cdot K_{a}}$$, thus $$x$$ can be ignored in the denominator:

$\mathrm{Solving\: for}\: x: \: \begin{matrix} \dfrac{x^{2}}{0.01}=10^{-5}\\ x^{2}=10^{-7} \\ x=3.2\times 10^{-4}=\mathrm{[H^{+}]} \end{matrix}$

(Note, compare this value with $$(0.01-x)$$ to see if neglecting $$x$$ in the equation for $$\mathrm{K_{a}}$$ was justified.)

$\mathrm{pH = -log[H^{+}]=-log(3.2\times 10^{-4})}$

$\mathrm{pH=3.2}$

Note: The degree of dissociation is the fraction of HA that dissociates.

$\mathrm{EQ5: \: degree\: of\: dissociation= \alpha = \dfrac{[A^{-}]}{[HA]_{init}}=\dfrac{[H^{+}]}{[HA]_{init}}}=\dfrac{[x]}{\mathrm{[HA]_{init}}}$

$\mathrm{For\: this\: problem,\: \alpha=\dfrac{3.2\times 10^{-4}}{0.01}=0.032}$

Thus, HA is 3.2% dissociated. The degree of dissociation depends on the concentration of HA. As the concentration of HA is decreased, $$\alpha$$ must increase in order to satisfy the equilibrium conditions.

### Class 2 Adding only the base ($$\mathrm{A^{-}}$$) form.

What is the pH of a $$0.01\: \mathrm{M}$$ solution of NaA? $$\mathrm{pK_{a}}$$ of HA= 5.0.

Solution:

NaA is a salt that completely dissociates in $$\mathrm{H_{2}O}$$ to give $$\mathrm{Na^{+}}$$ and $$\mathrm{A^{-}}$$.

This is a class 2 problem since only the conjugate base (base form) of HA is initially present. Since HA is a weak acid, the $$\mathrm{A^{-}}$$ will tend to combine with any available $$\mathrm{H^{+}}$$ to form HA. However, the only $$\mathrm{H^{+}}$$ available must come from $$\mathrm{H_{2}O}$$, which will result in the formation of $$\mathrm{OH^{-}}$$. The is the sum of two reactions:

$\mathrm{A^{-}+H^{+}\Leftrightarrow HA}$

$\mathrm{H_{2}O \Leftrightarrow H^{+} + OH^{-}}$

$\mathrm{EQ6:\: A^{-}+H_{2}O \Leftrightarrow HA+OH^{-}}$

A simple way to solve such problems uses a formula that is very similar to the formula used in Type 1 problems but which relates $$\mathrm{K_{b}}$$ and $$\mathrm{[OH^{-}]}$$.

First Calculate the $$\mathrm{pK_{b}}$$ from the $$\mathrm{pK_{a}}$$:

$\mathrm{EQ7: \: pK_{b}=14-pK{a}}$

Then calculate Kb from pKb

$\mathrm{EQ8: \: K_{b}=10^{-pK_{b}}}$

The $$\mathrm{[OH^{-}]}$$ can now be calculated directly from a simple formula similar to that used in Type 1 problems that derives from the definitions of $$\mathrm{K_{b}}$$:

$\mathrm{EQ9: \: K_{b}=\dfrac{[HA][OH^{-}]}{[A^{-}]}}$

Assuming that $$\mathrm{[OH^{-}]}=x$$ and $$C$$ = initial concentration of base added, then one obtains a formula nearly identical to that for a Class 1 problem:

$\mathrm{K_{b}=\dfrac{x^{2}}{c-x}}$

$$x$$ is then solved for exactly as in a Class 1 problem, and $$x=\mathrm{[OH^{-}]}$$. The pH is determined by the following simple calculations.

$\mathrm{pOH=-log[OH^{-}]}$

$\mathrm{pH=14-pOH}$

Using this method to solve the original problem:

$\mathrm{pK_{b}=14-pK_{a}=14-5.0=9.0}$

$\mathrm{K_{b}}=1\times 10^{-9}$

$1\times 10^{-9}=\dfrac{x^{2}}{c-x}=\dfrac{x^{2}}{0.01-x}$

ignore the “$$x$$” in the denominator because it should be $$<< 0.01$$:

$1\times 10^{-9}=\dfrac{x^{2}}{0.01}$

$x^{2}=1\times 10^{-11}$

$x=3.2\times 10^{-6}=\mathrm{[OH^{-}]}$

$\mathrm{pOH=-log}(3.2\times 10^{-6})=5.5$

$\mathrm{pH=14-pOH}=14-5.5=8.5$

The reaction of $$\mathrm{A^{-}}$$ with $$\mathrm{H_{2}O}$$ is often referred to as "hydrolysis". $$\mathrm{A^{-}}$$ in this case is acting as a base because its action as a proton acceptor leads to the production of $$\mathrm{OH^{-}}$$. Calculations identical to those above would be used for a base that directly releases $$\mathrm{OH^{-}}$$. That is, if BOH is a weak base (which can dissociate to form $$\mathrm{B^{+}}$$ and $$\mathrm{OH^{-}}$$) and $$\mathrm{B^{+}}$$ has a $$\mathrm{pK_{a}}$$ of 5.0 (which is the same as BOH having a $$\mathrm{pK_{b}}$$ of 9.0) then the calculation of the pH of $$0.01\: \mathrm{M}$$ solution of BOH would be identical to the above calculation. Only the concentration of the base and the pKa of the conjugate acid need be known to solve for pH. Similarly, if one knows the pH and the concentration of a base in a solution then the $$\mathrm{pK_{a}}$$ of the base can be determined. Or, the concentration can be calculated if the pH and the $$\mathrm{pK_{a}}$$ are known. Because of the easy mathematical relationship between the $$\mathrm{pK_{a}}$$ and the $$\mathrm{pK_{b}}$$, tables of these values usually list only the $$\mathrm{pK_{a}}$$ of the conjugate acid even if the compound is most commonly available in base form (such as for ammonia or Tris buffer).

### Class 3 Adding acid and base forms.

What is the pH of a solution containing $$0.02\: \mathrm{M}$$ HA and $$0.01\: \mathrm{M\: A^{-}}$$? $$\mathrm{pK_{a}}$$ of HA = 5.0.

Solution

Since this class of problem is the one most frequently encountered, and since pH rather than $$\mathrm{[H^{+}]}$$ is usually being calculated, a simplified method for solving these problems has been devised. This method still strictly depends on the definition of $$\mathrm{K_{a}}$$, but rearranges this definition into a more convenient form as follows:

$\mathrm{K_{a}=\dfrac{[H^{+}][A^{-}]}{[HA]}}$

$\mathrm{logK_{a}=log\dfrac{[H^{+}][A^{-}]}{[HA]}}$

$\mathrm{logK_{a}=log[H^{+}]+log\dfrac{[A^{-}]}{[HA]}}$

$\mathrm{-logK_{a}=-log[H^{+}]-log\dfrac{[A^{-}]}{[HA]}}$

$\mathrm{pK_{a}=pH-log\dfrac{[A^{-}]}{[HA]}}$

$\mathrm{pH=pK_{a}+log\dfrac{[A^{-}]}{[HA]}}$

This is the Henderson-Hasselbalch Equation.

Using this to solve the problem stated at the start of this section:

$\mathrm{pH=pK_{a}+log\dfrac{[A^{-}]}{[HA]}=5.0+log\dfrac{0.01\: M}{0.02\: M}}$

$\mathrm{pH=5.0+log(0.5)=4.7}$

### Sample Problem 1

a) A solution was prepared by dissolving 0.02 moles of acetic acid (HOAc; $$\mathrm{pK_{a}=4.8}$$) in water to give 1 liter of solution. What is the pH of this solution?

b) To this solution was then added 0.008 moles of concentrated sodium hydroxide (NaOH). What is the new pH? (In this problem, you may ignore changes in volume due to the addition of NaOH).

c) An additional 0.012 moles of NaOH is then added. What is the pH?

Solution

a) This is a class 1 problem: $$\mathrm{HOAc \Leftrightarrow H^{+}+OAc^{-}}$$

$$\mathrm{HOAc}$$ $$\mathrm{H^{+}}$$ $$\mathrm{OAc^{-}}$$
Initial conditions $$0.02\: \mathrm{M}$$ $$\sim 0$$ $$\sim 0$$
Change due to dissociation $$-x$$ $$+x$$ $$+x$$
Net concentration at equilibrium $$0.02-x$$ $$x$$ $$x$$

$\mathrm{K_{a}=\dfrac{[H^{+}][OAc^{-}]}{[HOAc]}=\dfrac{x^{2}}{0.02-x}=1.6\times 10^{-5}}$

$\mathrm{[HOAc]_{init}>100\times K_{a},\: so}\: x\: \mathrm{ignored\: in\: denominator}$

$\dfrac{x^{2}}{0.02}=1.6\times 10^{-5};x^{2}=3.2\times 10^{-7}$

$x=5.6\times 10^{-4}=\mathrm{[H^{+}]}$

$\mathrm{pH=-log(5.6\times 10^{-4})=3.25}$

b) NaOH is a strong base. It will react nearly quantitatively with HOAc to produce $$\mathrm{OAc^{-}}$$.

$\mathrm{HOAc+OH^{-}\Rightarrow OAc^{-}+H_{2}O}$

$0.008\: \mathrm{moles\: HOAc+0.008\: moles\: OH^{-}\Rightarrow 0.008\: moles\: OAc^{-}}$

$\mathrm{Amount\: of\: HOAc\: remaining=0.02-0.008=0.012\: moles.}$

$\mathrm{Thus, \: [HOAc]=0.012\: M\: and\: [OAc^{-}]=0.008\: M}$

Since both [HOAc] and $$\mathrm{[OAc^{-}]}$$ are present, this is a class 3 problem.

$\mathrm{pH=pK_{a}+log\dfrac{[OAc^{-}]}{[HOAc]}=4.8+log\dfrac{0.008\: M}{0.012\: M}}$

$\mathrm{pH=4.8+log(0.67)=4.8-0.17}$

$\mathrm{pH=4.63}$

c) After the addition of another 0.012 moles of NaOH, all of the HOAc has been converted to NaOAc. thus, $$\mathrm{[OAc^{-}]=0.02\: M}$$. Since only the "base form" of the HOAc is now present, this is a class 2 problem:

First find Kb:

$\mathrm{pK_{b}=14-[pK_{a}]=14-4.8=9.2}$

$\mathrm{K_{b}=10^{-pK_{b}}=10^{-9.2}=6.3\times 10^{-10}}$

$\mathrm{K_{b}=\dfrac{[HA][OH^{-}]}{[A^{-}]}}$

Setting $$\mathrm{[OH^{-}]=[HA]}=x$$

$\mathrm{K_{b}}=\dfrac{x^{2}}{\mathrm{[A^{-}]}-x}$

$6.3\times 10^{-10}=\dfrac{x^{2}}{0.02-x}\: \mathrm{can\: assume\:} x<<0.02$

$6.3\times 10^{-10}=\dfrac{x^{2}}{0.02}$

$x^{2}=1.26\times 10^{-11}$

$x=3.55\times 10^{-6}=\mathrm{[OH^{-}]}$

$\mathrm{pOH=-log(3.55\times 10^{-6})=5.45}$

$\mathrm{pH=14-pOH=14-5.45=8.55}$

Note: Every problem involving titration and buffers is similar to the sample problem just given. The addition of strong base (or acid) affects the initial concentrations of HA and $$\mathrm{A^{-}}$$. The pH can then be determined using the appropriate class 1, 2, or 3 solutions. Given the pH, the ratio of $$\mathrm{A^{-}}$$/HA can be determined from the Henderson-Hasselbalch Equation. The concentrations of $$\mathrm{A^{-}}$$ and HA can be determined if the total concentration ($$\mathrm{A^{-}+HA}$$) is known, or if the concentration of either species is known.

### Class 4 Polyprotic acids - amphoteric substances

Most situations that arise in problems involving polyprotic acids can be treated as class 1, 2, or 3 problems. There is one additional feature of polyprotic acids, the intermediate end points, that requires special treatment (Class 4).

Consider a polyprotic acid, $$\mathrm{H_{2}A}$$. It has two associated $$\mathrm{pK_{a}}$$ values as indicated below:

(\mathrm{pK_{a}}\) values as indicated below:

$\mathrm{H_{2}A \Leftrightarrow H^{+}+HA^{-}\: K_{a1},\: pK_{a1}}$

$\mathrm{HA^{-} \Leftrightarrow H^{+}+A^{2-}\: K_{a2},\: pK_{a2}}$

The following situations are possible:

$\mathrm{H_{2}A\: only\: \Rightarrow\: Class\: 1}$

$\mathrm{H_{2}A+HA^{-}\:\Rightarrow\: Class\: 3\: (use\: pK_{a1})}$

$\mathbf{HA^{-}\: only\: \Rightarrow\: Class\: 4}$

$\mathrm{HA^{-}+A^{2-}\: \Rightarrow\: Class\: 3\: (use\: pK_{a2})}$

$\mathrm{A^{2-}\: only\: \Rightarrow\: Class\: 2}$

“$$\mathrm{HA^{-}}$$ only" represents a special case because it is a weak acid itself and can undergo dissociation (Class 1) and it is also the base form of $$\mathrm{H_{2}A}$$ and can therefore undergo hydrolysis (Class 2). A good discussion of this system is provided in Segal, Biochemical Calculations, pp. 53-56.

A simplified way to look at it is to consider that when one molecule of $$\mathrm{HA^{-}}$$ dissociates to give an $$\mathrm{H^{+}}$$, another molecule of $$\mathrm{HA^{-}}$$ can pick it up to give $$\mathrm{H_{2}A}$$:

$\mathrm{HA^{-}\Leftrightarrow H^{+}+A^{2-}\: K_{eq}=K_{a2}}$

$\underline{\mathrm{H^{+}+HA^{-}\Leftrightarrow H_{2}A\: K_{eq}=1/K_{a1}}}$

$\mathrm{Net\: Rxn: \: 2HA^{-}\Leftrightarrow H_{2}A+A^{2-}\: K_{eq}=K_{a2}/K_{a1}}$

Note that as a result of this reaction the concentration of $$\mathrm{A^{2-}}$$ and $$\mathrm{H_{2}A}$$ will be equal. The pH of such a solution can be calculated by algebraically manipulating the Henderson-Hasselbalch Equation.

$\mathrm{1)\: pH=pK_{a1}+log\dfrac{[HA^{-}]}{[H_{2}]}}$

$\mathrm{2)\: pH=pK_{a2}+log\dfrac{[A^{2-}]}{[HA^{2}]}}$

$\mathrm{Sum:\: 2pH=pK_{a1}+pK_{a2}+log\dfrac{[HA^{-}][A^{2-}]}{[H_{2}A][HA^{-}]}}$

$\mathrm{2pH=pK_{a1}+pK_{a2}+log\dfrac{[A^{2-}]}{[H_{2}A]}}$

However, since $$\mathrm{[A^{2-}]=[H_{2}A]}$$, the last term is 0 and the equation simplifies to:

$\mathrm{2pH=pK_{a1}+pK_{a2}}$

$\mathrm{or}$

$\mathrm{pH=\dfrac{pK_{a1}+pK_{a2}}{2}}$

In general, for any intermediate salt of a polyprotic acid, the pH will be the average of the two $$\mathrm{pK_{a}}$$ values that are associated with that particular ionic species.