Acid-Base Chemistry

pH of Strong Acid

\(\ce{HCl_{(aq)} -> H+_{(aq)} + Cl-_{(aq)}}\)

A strong acid completely dissociates in solution. As such, a single reaction arrow is used to indicate that the reaction is almost 100% complete at equilibrium. To find the pH of a strong acid, take the -log of the hydrogen ion concentration [H⁺]. For example, what is the pH of a 0.5 M solution of HCl? $$\text{pH} = -\log_{10}(0.5\, \text{M}) = 0.301$$

pH of Weak Acid

\(\ce{CH3COOH_{(aq)} <=> CH3COO^-_{(aq)} + H^+_{(aq)}}\)

A weak acid does not completely dissociate in solution. The degree of dissociation (and thus strength of the acid) is determined by the acid dissociation constant (Ka). The Ka is the equilibrium constant for the dissociation of the acid:

\(\ce{K_a = \frac{{[H+][A-]}}{{[HA]}}}\)

What is the pH of a 0.1 M aqueous solution of acetic acid? To solve this problem, we must first establish the Ka expression for acetic acid:

\(K_a = \frac{[\ce{CH3COO^-}][\ce{H^+}]}{[\ce{CH3COOH}]}\)

An ICE table (Initial, Change, Equilibrium) is used to determine the pH at equilibrium based on initial conditions.

Reaction:	\( \ce{CH3COOH} \)	\( \rightleftharpoons \)	\( \ce{H+} \)	\( + \)	\( \ce{CH3COO-} \)
Initial (M):	\( 0.1 \)		\( 0 \)		\( 0 \)
Change (M):	\( -x \)		\( +x \)		\( +x \)
Equilibrium (M):	\( 0.1 - x \)		\( x \)		\( x \)

Next, we substitute these variables into our original equilibrium expression:

\( K_a = \frac{(x)(x)}{0.1 - x} \)

Assuming that x is small relative to the concentration of acid, we can ignore when it is subtracted from 0.1 M in the equilibrium expression. Next, we solve for x, which reflects the [H⁺] at equibrium. We then take the -log of this value to get the pH. The Ka of acetic acid is 1.8x10^-5. Solving for x we get: \begin{align*} & 1.8 \times 10^{-5} = \frac{x^2}{0.1} \\ & x = \sqrt{1.8 \times 10^{-5} \times 0.1} = 1.34 \times 10^{-3} \\ & \text{pH} = -\log[H^+] = -\log(1.34 \times 10^{-3}) = 2.87 \\ \end{align*}

pH of Strong Base

Calculating the pH of a strong base is similar to that of a strong acid. A strong base is assumed to completely dissociate in solution. Consider the reaction of the strong base sodium hydroxide in water:

\(\ce{NaOH (s) -> Na+ (aq) + OH- (aq)}\)

To determine the pH, we take the -log of the [OH^-], and then subtract this from 14. For example, what is the pH of a 1 M solution of NaOH?

\(\ce{pH = 14 - pOH = 14 - (-log_{10}(1)) = 14}\)

pH of Weak Base

Determining the pH of a weak base is similar to that of a weak acid. We must first establish an equilibrium expression for the weak base. A weak base does not fully dissociate in solution. The degree to which it dissociates, and thus its strength, is determined by the base dissociation constant, Kb. Consider the following weak base, ammonia, in solution:

\(\ce{NH3 (aq) + H2O (l) -> NH4+ (aq) + OH- (aq)}\)

The equilibrium expression for this reaction is:

\(\ce{Kb = \frac{[NH4+][OH-]}{[NH3]}}\)

Let's consider the following example: What is the pH of a 0.2 M solution of NH3?

As we did with the weak acid, we can setup an ICE table to determine the conditions at equilibrium:

Reaction:	\( \ce{NH3} \)	\( \rightleftharpoons \)	\( \ce{NH4+} \)	\( + \)	\( \ce{OH-} \)
Initial (M):	\( 0.2 \)		\( 0 \)		\( 0 \)
Change (M):	\( -x \)		\( +x \)		\( +x \)
Equilibrium (M):	\( 0.2 - x \)		\( x \)		\( x \)

Substituting these values into our equilibrium expression we get:

\(\ce{Kb = \frac{(x)(x)}{0.2 - x}}\)

Since x is small relative to the concentration of ammonia, we can ignore the subtraction of x in the denominator. The Kb of ammonia is 1.8x10^-5. Solving for x, and thus the [OH^-], we get:

\(\ce{x = \sqrt{1.8 \times 10^{-5} \times 0.2} = 0.0019}\)

To find the pH, we take -log(0.0019) to get a pOH of 2.72, 14 - 2.72 = 11.28.

Acid-Base Titrations

An acid-base titration is used to determine the unknown concentration of an acid or base. The process involves gradually adding a titrant of known concentration to an analyte of known volume but unknown concentration. The point at which the titrant has completely neutralized the analyte is known as the equivalence point. At this point, the moles of acid are equal to the moles of base, and this information can be used to determine the concentration of the analyte.

By graphing the pH as a function of volume of titrant added, a titration curve can be generated. Various points on the titration curve are of interest with regard to the specific chemical species present at that time and resulting pH. Consider the following titration of 50 mL of 0.1 M HCl with 0.1 M NaOH:

There are a few key points to assess on this titration curve:

Initial pH: The pH of the solution before any NaOH is added is determined solely by the HCl present. Since HCl is a strong acid, it can be assumed to dissociate completely in solution. Take the negative base 10 log of the hydrogen ion concentration to get the pH:

\[ \text{pH} = -\log_{10}[\ce{H+}] = -\log_{10}(0.1) = 1 \]

pH Between the Initial pH and the Equivalence Point: At this point during the titration, the added NaOH reacts with HCl as follows:

\(\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}\)

The stoichiometry of this reaction will determine the pH. HCl and NaOH react in equimolar amounts during this reaction, which results in the production of a neutral salt (NaCl) and water. Therefore, the pH at this point will be determined by the moles of HCl remaining following the reaction. When calculating the resulting molarity, it is important to make sure that calculation reflects that increase in total volume from the added NaOH.

For example, let's say we want to calculate the pH after 10 mL of NaOH has been added. First, let's determine how many moles of NaOH this represents. Since the concentration of NaOH is 0.1 M, we can multiply that by the volume to get total moles:

\(\ce{0.1 M NaOH} \times 0.010 \, \text{L} = 0.001 \, \text{moles NaOH}\)

Similarly, we can find the total moles of HCl present from the start of the titration as follows:

\(\ce{0.1 M HCl} \times 0.050 \, \text{L} = 0.005 \, \text{moles HCl}\)

0.001 moles of NaOH will react with the 0.001 moles of HCl, leaving 0.005 - 0.001 = 0.004 moles of HCl in solution, which will determine the pH. The new total volume is the original 50 mL of HCl + 10 mL NaOH added = 60 mL. The new molarity of HCl and resulting [H⁺] in solution is:

\(\frac{0.004 \, \text{moles} \, \ce{HCl}}{0.060 \, \text{L}} = 0.067 \, \text{M} \, \ce{HCl}\)

From here, we can calculate the pH as follows:

\(\text{pH} = -\log_{10}(0.067) = 1.17\)

The pH of the solution after the addition of 10 mL of NaOH has risen to 1.17. The pH at other points between the initial pH and equivalence point can be calculated in a similar fashion.

pH at the Equivalence Point: The equivalence point is the point at which the analyte has completely reacted with the titrant. At this point, the moles of acid are equal to the moles of base. A pH indicator (halochromic chemical compound that changes color based on pH) is useful in determining the volume of titrant added at which the equivalence point occurs. Again, this is a fundamental aspect of titrations: given a known volume but unknown concentration of analyte, knowing the volume of titrant necessary to reach the equivalence point can be used to determine the unknown concentration.

In the above example, the equivalence point occurs when 50 mL of NaOH has been added. At this point, all .005 moles of NaOH have reacted with all 0.005 moles of HCl. The only species present in aqueous solution at this point is the neutral salt NaCl. Thus, the pH at the equivalence point of a strong acid-strong base titration is neutral.

pH Past the Equivalence Point: Past the equivalence point, NaOH base is being added in excess, resulting in a further increase in pH. To calculate the pH at a point past the equivalence point, we must determine how much OH^- exists in excess, from which we can calculate the pOH and then pH.

For example, what is the pH after 60 mL of NaOH has been added. After the addition of 50 mL, the HCl reacted fully with the NaOH. Thus, the excess volume of NaOH at 60 mL is 60 mL - 10 mL = 10 mL. This represents 10 mL * 0.1 M = 0.001 moles of NaOH. Since NaOH is a strong base, it dissociates fully in solution. Again, we must make sure we consider the total volume when determining the molarity of NaOH in solution. The total volume at this point is 110 mL (50 mL of HCl starting + 60 mL of NaOH added), so the molarity of NaOH is 0.001 moles/110 mL = 0.009 M. The pOH and then pH can be determined as follows:

\(pOH = -\log_{10}[\ce{OH-}] = -\log_{10}(0.009) = 2.05\)
\(pH = 14 - pOH = 14 - 2.05 = 11.95\)

Strong Acid-Strong Base Titrations

Let's review the key features of a strong acid-strong base titration, of which the above example of the titration of HCl with NaOH is an example. Of note:

• The pH at the start of the titration depends solely on the strong acid.
• The pH between the start of the titration and the equivalence point can be determined stoichiometrically by calculating the number of moles of strong acid that have reacted with strong base. It is important to account for the total volume as strong base is added when calculating the resulting molarity.
• The pH at the equivalence point is neutral (7).
• The pH past the equivalence point can be determined by calculating the moles of excess base, determining the pOH, and then determining the pH.

Weak Acid-Strong Base Titrations

A weak acid-strong base titration differs from a strong acid-strong base titration due to the presence of a weak instead of strong acid. A weak acid, by definition, will not dissociate fully in solution. Instead, the degree of dissociation and thus strength of the acid is determined by its acid dissociation constant (Ka). The Ka and associated equilibrium expression are used to calculate the pH of weak acids in solution (see prior section on calculating pH of a weak acid). Some additional key differences between a strong acid-strong base titration and a weak acid-strong base are:

• For a given weak acid and strong acid of the same molarity, the weak acid has a higher starting pH than the strong acid.
• The pH at the equivalence point is greater than 7 due to the predominance of a basic species at this point.
• The dynamics of the titration curve are more gradual in the region of the equivalence point as compared to the strong acid.
• A buffer region exists between the starting point and the equivalence point due to the presence of a weak acid-base conjugate pair.

Let's consider the titration of the weak acid acetic acid (CH3COOH) with the strong base sodium hydroxide (NaOH):



Let's consider the key points and associated calculations on this titration curve:

Starting Point: At the start of the titration, no titrant (NaOH) has been added. As such, the pH is entirely dependent on the weak acid, CH3COOH. This is analogous to the weak acid calculation above. The starting pH is 2.87.

Between the Starting Point and Equivalence Point: As NaOH is added to the analyte, the following reaction occurs:

\(\ce{CH3COOH(aq) + NaOH(aq) -> CH3COONa(aq) + H2O(l)}\)

Sodium acetate (CH3COONa) is a strong electrolyte, and as such will dissociate completely in solution to Na⁺ and acetate (CH3COO^-). The sodium ion does not have acid or base properties, but the acetate is the conjugate base of acetic acid. Hence, we have a weak acid-base pair (acetic acid and acetate), forming a buffer solution. The pH can be calculated here using the Henderson-Hasselbalch equation.

\(\ce{pH = pKa + \log\left(\frac{[A^-]}{[HA]}\right)}\)

For example, what is the pH when 10 mL of NaOH has been added? First, let's determine how many moles of NaOH this corresponds to by multiplying by the molarity (0.1 M) to get 0.001 moles. These 0.001 moles of NaOH will react with CH3COOH, given a final amount of CH3COOH equal to 0.005 moles - 0.0001 moles = 0.004 moles. Based on the above reaction, 0.001 moles of sodium acetate, and thus ultimately acetate in solution, will be produced.

Next, we must consider the concentration of species in solution. What is our total volume? We started with 50 mL of acetic acid and added 10 mL of NaOH, so 60 mL. Using this volume, we find the molarity of acetic acid to be (0.004 moles/60 mL) = 0.067 M and the molarity of acetate to be (0.001/60 mL) = 0.0167. Substituting these values into the Henderson-Hasselbalch equation, along with the pKa of acetic acid (4.76) gives us:

\(\ce{pH = 4.76 + \log_{10}\left(\frac{[CH3COO^-]}{[CH3COOH]}\right) = 4.76 + \log_{10}\left(\frac{0.0167}{0.067}\right) = 4.16}\)

Half-Equivalence Point
What is unique about the first point labeled on the graph? At this point, half of the volume of NaOH has been added relative to the equivalence point. In solution at this point, the concentration of acetic acid is equal to acetate, and thus the Henderson-Hasselbalch equation becomes:

\(\ce{pH = pKa}\)

This is known as the half-equivalence point.

Equivalence Point: At the equivalence point, NaOH has fully reacted with the acetic acid. As such, the only species present in solution is sodium acetate (CH3COONa), a strong electrolyte. The acetate ion is a weak base, and is responsible for the pH at the equivalence point. Based on the stoichiometry of the above reaction, at the equivalence point the moles of sodium acetate, and thus acetate ion, is equal to the moles of NaOH added. 50 mL of .1 M NaOH were added = 0.005 moles. There are 0.005 moles to acetate ion present in solution at this point, in a total volume of (50 mL initial acetic acid + 50 mL NaOH added) 100 mL, resulting in a concentration [CH3COO^-] = 0.05 M. This is weak base calculation that can be solved with an ICE table and Kb, similar to the prior calculation for a weak acid. At equilibrium, the expression for acetate is:

\(\ce{Kb = \frac{[CH3COOH][OH^-]}{[CH3COO^-]}}\)

The Kb of acetate is 5.6 x 10^-10. Algebraically, the expression is:

\(\ce{Kb = \frac{x \cdot x}{[CH3COO^-] - x}}\)

Since the concentration of acetate is large relative to x, we can ignore the subtraction of x in the denominator. Solving for x gives us the concentration of hydroxide:

\(\ce{[OH^-] = \sqrt{0.05 \times 5.6 \times 10^{-10}} = 5.29 \times 10^{-6}}\)

\(\ce{pH = 14 - pOH = 14 - (-\log_{10}(5.29 \times 10^{-6})) = 8.7}\)

Thus, the pH at the equivalence point is basic at 8.7.

pH Past the Equivalence Point: Past the equivalence point, acetate and NaOH will be present in solution. However, NaOH is a significantly stronger base than acetate, such that at this point the contribution to pH from acetate will be negligible. To solve for pH here, we determine the moles of excess base present followed by the molarity of hydroxide in solution. For example, what is the pH when 75 mL of NaOH has been added?

At this point, there are 75 mL * 0.1 M = 0.0075 moles of NaOH added that react with 0.05 moles of CH3COOH resulting in 0.0075 - 0.005 = 0.0025 moles excess. What is the total volume at this point? 50 mL of acetic acid + 75 mL NaOH added = 125 mL, giving a concentration of NaOH of 0.0025 moles/0.125 L = 0.020 M. Since NaOH is a strong base it dissociates in solution, the pH = 14 - pOH = 14 - 1.7 = 12.3.

Strong Base-Strong Acid Titrations

Let's consider the titration of 0.1 M NaOH with 0.1 M HCl:

This curve has the same S-shape as the strong acid-strong base titration but is inverted. The calculations are similar to the strong acid-strong base titration, except one starts with a strong base instead of a strong acid. The initial pH is determined entirely by the NaOH analyte. As HCl is added, the pH is determined by the moles of NaOH present after reacting with HCl followed by determining the molarity of NaOH (again, it is important to keep track of total volume). At the equivalence, the pH is neutral as the neutral salt NaCl is the only species present. As HCl continues to be added, moles of excess HCl divided by the total volume determine the pH.

Weak Base-Strong Acid Titrations

Consider the titration of the weak base 0.1 M NH3 with the strong acid 0.1 M HCl:

Let's examine the four key points on this titration curve:

Starting Point: The pH at the starting point is determined entirely by the weak base NH3. This is the same as the prior weak base calculation.

Between the Starting Point and the Equivalence Point: The following chemical reaction takes place as NCl is added to the analyte:

\(\ce{NH3(aq) + HCl(aq) -> NH4Cl(aq)}\)

Ammonium chloride is a strong electrolyte and will dissociate completely in solution:

\(\ce{NH4Cl(aq) -> NH4^+(aq) + Cl^-(aq)}\)

So what do we have at this point of the titration? As HCl is added and reacts with NH3, we form ammonium chloride, which then dissociates into the ammonium and chloride ions. Ammonium is the conjugate acid of ammonia. We have a conjugate acid-base pair, or buffer. Thus, the Henderson-Hasselbach equation can be used to calculate the pH in this region of the titration:

\(\text{pH} = \text{pKa} + \log{\left(\frac{[\ce{NH3}]}{[\ce{NH4+}]}\right)} \)

The Ka for ammonium is 5.6x10^-10. For example, what is the pH when 25 mL of 0.1 M HCl has been added? At this point, 0.1 M * 25 mL = 0.0025 moles of HCl react with the 0.005 moles of NH3 ily present, leaving 0.0025 moles of NH3 and producing 0.0025 moles of NH4⁺ by the above reaction. What is the total volume? 50 mL of NH3 starting + 25 mL HCl added = 75 mL. For here, we find the molarities of NH3 and NH4⁺ to both 0.033 M. Thus, the pH by the Henderson-Hasselbach equation is:

\(\text{pH} = 9.25 + \log{\left(\frac{0.033}{0.033}\right)} = 9.25\)

What is unique about this point in the tiration? This is again the half-equivalence point (similar to what we saw previously with a weak acid-strong base titration), in which the pH = the pKa. In this case, the pKa of ammonium is 9.25, which is the pH when 25 mL of HCl have been added.

Equivalence Point: At the equivalence point, all of the starting ammonia has been consumed by the reaction. We are left with ammonium, a weak acid. The moles of ammonium present at the equivalence point are equal to the moles of ammonia we started with. The total volume here is 100 mL. Using this information, we can calculate the pH of this weak acid using an ICE table (similar to what we have done prior). Since ammonium is an acidic species, the pH at the equivalence point is acidic.

pH Past the Equivalence Point: Past the equivalence point, ammonium and excess HCl will be present in solution. Since HCl is a significantly stronger acid than ammonium, the pH will be determined by the concentration of HCl at this point. This is a strong acid calculation.