# Buffers II

## Titration of Weak Acid by a Strong Base

Skills to Develop

• Calculate the pH and plot the titration curve when a weak acid is titrated by a strong base.
• Solve buffer problems.
• Solve hydration (hydrolyis> problems.
• Prepare a buffer of desirable pH.

Buffer solutions contain a weak acid and its salt or a weak base and its salt. The pH values of these solutions do not change much when a little bit of acid or base is added. On this page, we explore the reasons why the pH of buffer solutions resists change. The blood is a natural buffer, and so are other body fluids and plant fluids due to mixtures of weak acids and bases present in them. Buffer solutions are required for many chemical experiments; they are also useful to standardize pH meters.

You have investigated how the pH varies in a strong-acid and strong-base Titration. We illustrate the titration of a weak acid by a strong base using the following examples. During the titration process and before the equivalence point is reached, some acid has been neutralized by the strong base, and the solution contains a weak acid and its salt. The solution acts as a buffer.

Example $$\PageIndex{1}$$

What is the pH of a Ca M acid solution whose acid dissociation constant is Ka? (What is the pH of a Ca M weak acid before starting the titration?)

SOLUTION

Let $$\ce{HA}$$ represent the weak acid, and assume x M of it is ionized. Then, the ionization and equilibrium concentration is

$$\begin{array}{ccccc} \ce{HA &\rightleftharpoons &H+ &+ &A-}\\ C_{\large\textrm a}-x &&x &&x \end{array}$$

$$K_{\large\textrm a} = \dfrac{x^2}{C_{\large\textrm a}-x}$$

$$x^2 + K_{\large\textrm a}x - C_{\large\textrm a}K_{\large\textrm a} = 0$$

$$x = \dfrac{-K_{\large\textrm a} + (K_{\large\textrm a}^2 + 4 C_{\large\textrm a}K_{\large\textrm a})^{1/2}}{2}$$

$$\ce{pH} = -\log(x)$$

DISCUSSION

The method has been fully discussed in Weak Acids and Bases equilibrium. Symbols are used here, but approximations may be applied to numerical problems.

Example $$\PageIndex{2}$$

Let us make a buffer solution by mixing Va mL of acid $$\ce{HA}$$ and Vs mL of its salt $$\ce{NaA}$$. For simplicity, let us assume both the acid and the salt solutions have the same concentration C M. What is the pH of the so prepared buffer solution? The acid dissociation constant is Ka.

SOLUTION

After mixing, the concentrations Ca and Cs of the acid $$\ce{HA}$$ and its salt $$\ce{NaA}$$ respectively are

$$C_{\large\textrm a} = \dfrac{C\, V_{\large\textrm a}}{V_{\large\textrm a}+V_{\large\textrm s}}$$

$$C_{\large\textrm s} = \dfrac{C\, V_{\large\textrm s}}{V_{\large\textrm a}+V_{\large\textrm s}}$$

Assume x M of the acid is ionized. Then, the ionization and equilibrium of the acid is shown below, but the salt is completely dissociated.

$$\begin{array}{ccccc} \ce{HA &\rightleftharpoons &H+ &+ &A-}\\ C_{\large\textrm a}-x &&x &&x \end{array}$$

$$\begin{array}{ccccc} \ce{NaA &\rightarrow &Na+ &+ &A-}\\ &&C_{\large\textrm s} &&C_{\large\textrm s} \end{array}$$

$$\ce{Common\: ion\: [A- ]} = x+C_{\large\textrm s}$$

$$K_{\large\textrm a} = \dfrac{x(x+C_{\large\textrm s})}{C_{\large\textrm a}-x}$$

$$x^2 + (K_{\large\textrm a}+C_{\large\textrm s})x - C_{\large\textrm a} K_{\large\textrm a} = 0$$

$$x = \dfrac{ -(K_{\large\textrm a}+C_{\large\textrm s}) + ((K_{\large\textrm a}+C_{\large\textrm s})^2 + 4 C_{\large\textrm a} K_{\large\textrm a})^{1/2}}{2}$$

$$\ce{pH} = -\log(x)$$

DISCUSSION

The formulas for x and the pH derived above can be used to estimate the pH of any buffer solution, regardless how little salt or acid is used compared to their counterpart.

When the ratio Ca / Cs is between 0.1 and 10, the Henderson-Hasselbalch equation is a convenient formula to use.

$$K_{\large\textrm a} = \ce{\dfrac{ [H+] [A- ]}{[HA]}}$$

$$\mathrm{p\mathit K_{\large\textrm a} = pH} - \log \left(\dfrac{\ce{[A- ]}}{\ce{[HA]}}\right)$$

The Henderson-Hasselbalch equation is

\begin{align} \ce{pH} &= \mathrm{p\mathit K_{\large\textrm a}} + \log \left(\dfrac{\ce{[A- ]}}{\ce{[HA]}}\right)\\ &= \mathrm{p\mathit K_{\large\textrm a}} + \log \left(\dfrac{ [C_{\large\textrm s}] + x}{[C_{\large\textrm a}] - x}\right) \end{align}

Because when x is insignificant in comparison to Cs, $$\ce{[A- ]} = C_{\large\textrm s}$$ and $$\ce{[HA]} = C_{\large\textrm a}$$.
 Base [H+] pH added See 0.0 mL 0.00316 2.500 A. 0.1 9.07e-4 3.042 B. 1.0 9.07e-5 4.042 C. 5.0 1.0e-5 5.000 D. 10.0 10-11.349 11.349 E.

Example $$\PageIndex{3}$$

Plot the titration curve when a 10.00 mL sample of 1.00 M weak acid $$\ce{HA}$$ (Ka = 1.0e-5) is titrated with 1.00 M $$\ce{NaOH}$$.

SOLUTION

1. Because the concentration is high, we use the approximation

\begin{align} \ce{[H+]} &= (C_{\large\textrm a}K_{\large\textrm a})^{1/2}\\ &= 0.00316\\ \ce{pH} &= 2.500 \end{align}

Note the sharp increase in pH when 0.1 mL (3 drops) of basic solution is added to the solution.

2. When 0.1 mL $$\ce{NaOH}$$ is added, the concentration of salt (Cs), and concentration of acid (Ca) are:

$$C_{\large\textrm s} = \mathrm{\dfrac{0.1\times1.0\: M}{10.1} = 0.0099\: M}$$

$$C_{\large\textrm a} = \mathrm{\dfrac{9.9\times1.0\: M}{10.1}} = 0.98$$

$$\begin{array}{ccccc} \ce{HA &\rightleftharpoons &H+ &+ &A-}\\ C_{\large\textrm a}-x &&x &&x \end{array}$$

$$\ce{[A- ]} = x + 0.0099\: (= C_{\large\textrm s})$$

$$K_{\large\textrm a} = \dfrac{ x\, (x + 0.0099)}{0.98 - x} = \textrm{1e-5}$$

$$x^2 + 0.0099 x = 9.8\textrm{e-}6 - 1\ce e5 x$$

$$x^2 + (0.0099 + 1\ce e5)\, x - 9.8\textrm{e-}6 = 0$$

\begin{align} x &= \dfrac{-0.0099 + (0.00992 + 4\times 0.98\times \textrm{1e-5})^{1/2}}{2}\\ &= 0.000907 \end{align}

$$\ce{pH} = 3.042$$

Note that using the Henderson-Hasselbalch equation will not yield the correct solution. Do you know why?

3. When 1.0 mL $$\ce{NaOH}$$ is added,

$$C_{\large\textrm s} = \mathrm{\dfrac{1.0\times1.0\: M}{11.0} = 0.0901\: M}$$

$$C_{\large\textrm a} = \mathrm{\dfrac{9.0\times1.0\: M}{11.0} = 0.818}$$

$$\begin{array}{ccccc} \ce{HA &\rightleftharpoons &H+ &+ &A-}\\ C_{\large\textrm a}-x &&x &&x+0.0901 \end{array}$$

$$K_{\large\textrm a} = \dfrac{x\, (x + 0.0901)}{0.818 - x} = \textrm{1e-5}$$

\begin{align} x &= \dfrac{-0.0901 + (0.09012 + 4\times0.818\times1\textrm{e-}5)^{1/2}}{2}\\ &= \textrm{0.0000907 (any approximation to be made?)} \end{align}

$$\mathrm{pH = 4.042}$$

Using the Henderson Hasselbalch approximation yields

$$\mathrm{pH = p\mathit K_{\large\textrm a}} + \log \left(\dfrac{ 0.0901(=\ce{[A- ]})}{0.818(=\ce{[HA]})}\right) = 5 - 0.958 = \textrm{4.042 (same result)}$$

4. When 5.0 mL $$\ce{NaOH}$$ is added,

$$C_{\large\textrm s} = \mathrm{\dfrac{5.0\times1.0\: M}{15.0} = 0.333\: M}$$

$$C_{\large\textrm a} = \mathrm{\dfrac{5.0\times1.0\: M}{15.0} = 0.333\: M}$$

$$\begin{array}{ccccc} \ce{HA &\rightleftharpoons &H+ &+ &A-}\\ C_{\large\textrm a}-x &&x &&x+0.3333 \end{array}$$

$$K_{\large\textrm a} = \dfrac{x\, (x + 0.333)}{0.333 - x} = \textrm{1.0e-5}$$

$$x^2 + (0.333-1\textrm{e-}5)\,x - 0.333\times\textrm{1e-5} = 0$$

\begin{align} x &= \dfrac{-0.333 + (0.3332 + 4\times3.33\textrm{e-}6)^{1/2}}{2}\\ &= 0.0000010 \end{align}

$$\mathrm{pH = 5.000}$$

Note: Using the Henderson Hasselbalch approximation yields the same result

$$\mathrm{pH = p\mathit K_{\large\textrm a}} + \log \left(\dfrac{0.333(=\ce{[A- ]})}{0.333(=\ce{[HA]})}\right) = 5 + 0.000 = 5.000$$

5. When 10.0 mL $$\ce{NaOH}$$ is added,

$$C_{\large\textrm s} = \mathrm{\dfrac{10.0\times1.0\: M}{20.0} = 0.500\: M}$$

$$C_{\large\textrm a} = \mathrm{\dfrac{0.0\times1.0\: M}{20.0} = 0.000}$$

At the equivalence point, the solution contains 0.500 M of the salt $$\ce{NaA}$$, and the following equilibrium must be considered:

$$\begin{array}{cccccccl} \ce{A- &+ &H2O &\rightleftharpoons &HA &+ &OH- &}\\ C_{\large\textrm s}-x &&&&x &&x &\mathrm{Equilibrium\:\, concentrations} \end{array}$$

\begin{align} K_{\large\textrm b} &= \dfrac{\ce{[HA] [OH- ]}}{\ce{[A- ]}} \dfrac{\ce{[H+]}}{\ce{[H+]}}\\ &= \dfrac{K_{\large\textrm w}}{K_{\large\textrm a}} = \dfrac{\textrm{1e-14}}{\textrm{1e-5}} = \textrm{1e-9} = \dfrac{x^2}{C_{\large\textrm s}-x} \end{align}

$$x = (0.500\times\textrm{1.0e-9})^{1/2} = \textrm{2.26e-5}$$

$$\textrm{pOH} = -\log x = 2.651;\: \textrm{pH} = 14 - 2.651 = 11.349$$

Note: The calculation here illustrates the hydration or hydrolysis of the basic salt $$\ce{NaA}$$.

DISCUSSION

Sketch the titration curve based on the estimates given in this example, and notice the points made along the way.

## Questions

1. Plot the titration curve when a 10.00 mL sample of 0.100 M weak acid $$\ce{HA}$$ (Ka = 10e-6) is titrated by 0.100 M $$\ce{NaOH}$$ solution.
2. The compound $$\ce{HF}$$ is a weak acid, Ka = 6.7e-4. Calculate Kb and pKb for the fluoride ion $$\ce{F-}$$ in an aqueous solution.
3. During the titration of 10.0 mL 0.10 M $$\ce{HF}$$ solution with a 0.10 M $$\ce{NaOH}$$ solution, 1.0 mL of $$\ce{NaOH}$$ has been added. What is the concentrations of sodium ion $$\ce{[Na+]}$$ and fluoride ion $$\ce{[F- ]}$$?
4. During the titration of 10.0 mL 0.10 M $$\ce{HF}$$ solution with a 0.10 M $$\ce{NaOH}$$ solution, 5.0 mL of $$\ce{NaOH}$$ has been added. What is the concentrations of sodium ion $$\ce{[Na+]}$$ and fluoride ion $$\ce{[F- ]}$$?
5. During the titration of 10.0 mL 0.10 M $$\ce{HF}$$ (Ka = 6.7e-4) solution with a 0.10 M $$\ce{NaOH}$$ solution, 5.0 mL of $$\ce{NaOH}$$ has been added. What is the pH at this point?
6. When equal volumes of 0.10 M $$\ce{HF}$$ and 0.10 M $$\ce{NaOH}$$ are mixed, what is the pH of such a solution?

## Solutions

1. Hint..
Learn it by doing! Perform all the calculations outlined in Example 3, and then you will be an expert for solving buffer and hydration problems.

2. Hint..
The following relationship is handy to use, but make sure you know why.

$$K_{\large\textrm b} = \dfrac{K_{\large\textrm w}}{K_{\large\textrm a}} =\: ?$$

What is the pH at the equivalence point when a 0.10 M $$\ce{HF}$$ solution is titrated by a 0.10 M $$\ce{NaOH}$$ solution?

3. Hint..
A simple consideration leads to the formulation:

$$\ce{[Na+]} = \ce{[F- ]} = \mathrm{\dfrac{1.0\: mL \times 0.10\: M}{11.0\: mL}}$$

4. Answer $$\ce{[F- ]} = \mathrm{0.0333\: M}$$

5. Answer pH = pKa = 3.17

6. Hint..

What is the pH of the equivalence point during the titration of 0.10 M $$\ce{HF}$$ using 0.10 M $$\ce{NaOH}$$ solution? (Another way of asking the same question.)