How to use equivalence point to find concentration

Example 1

You are given 90 mL of 0.6 M of the weak base NH3 (Kb = 1.8 × 10-5), and 1 M of the strong acid titrant HCl.

1. What is the pH before any acid is added?
2. What volume of acid (in mL) is needed to reach the equivalence (stoichiometric) point?
3. What volume of acid (in mL) is needed to reach the halfway point where pH = pKa?
4. What is the pH after 50 mL of acid is added?
5. What is the pH at the equivalence (stoichiometric) point?
6. What is the pH after 60 mL of acid is added?

First, calculate the number of moles of base (analyte) present initially.

\[0.090 \; L \; base \; \times \dfrac{0.6 \; mol \; base}{L \; base \; solution} = 0.054 \; mol \; base\]

a) An ICE table helps determine the molarity of OH-.

\[1.8 \times 10^{-5} = \dfrac{x^2}{0.6 - x}\]

\[1.08 \times 10^{-5} - 1.8 \times 10^{-5x} - x^2 = 0\]

\[x = \dfrac{1.8 \times 10^{-5} \pm \sqrt{(1.8 \times 10^{-5})^2 - 4(-1)(1.08 \times 10^{-5})}}{2(-1)}\]

\(= \dfrac{1.8 \times 10^{-5} \pm 6.57 \times 10^{-3}}{-2} = -3.29 \times 10^{-3}, \; 3.28 \times 10^{-3} \; M \; OH^-\)

\[pOH = -log(3.28 \times 10^{-3}) = 2.5 \; pOH\]

\[pH = 14 - pOH = 14 - 2.5 = 11.5 \; pH\]

b) At the equivalence point, the number of moles of HCl added is equal to the initial number of moles of NH3, because the analyte is completely neutralized.

\[0.054 \; mol \; HCl \times \dfrac{1 \; L \; HCl}{mol \; HCl} = 0.054 \; L \; HCl, \; or \; 54 \; mL \; HCl\]

c) At the midpoint, pOH = pKb.

\(-log(1.8 \times 10^{-5}) = 4.74 \; pOH\)

\(pH = 14 - pOH = 14 - 4.74 = 9.26 \; pH\)

At the midpoint, the number of moles of HCl added equals half the initial number of moles of NH3. In other words, the number of moles of HCl added at the midpoint is half of the number of moles of HCl added by the equivalence point. Hence:

\[\dfrac{1}{2}(0.054 \; L \; HCl \; added \; at \; equivalence \; point) = 0.027 \; moles \; HCl \; at \; midpoint \]

\[\text{Volume of acid needed} = 0.027\ mol\ HCl \times \dfrac{1\ L}{1\ mol\ HCl} = 0.027\ L\ HCl = 27\ mL\ HCl \]

d) First, find the moles of HCl in 50 mL of HCl.

\[0.05 \; L \; HCl \times \dfrac{mol \; HCl}{L \; HCl} = 0.05 \; mol \; HCl\]

Because 50 mL of acid have been added, and we started out with 90 mL of analyte, there are a total of 140 mL of analyte solution at this point. Hence, the molarity of NH3 is the following:

\(\dfrac{0.004 \; mol \; NH_3}{0.140 \; L \; solution \; in \; flask}=0.0286 M\)

The molarity of NH4+ is:

\(\dfrac{0.050 \; mol \; NH_4^{+}}{0.140 \; L \; solution \; in \; flask}=0.357 M\)

Now we can use the Henderson-Hasselbalch approximation:

\(pOH=pK_b+log\dfrac{NH_4^{+}}{NH_3}\)

\(pOH=4.74+log\dfrac{0.357}{0.0286}=5.84 \; pOH\)

\(pH=14-pOH=14-4.84=8.16\)

e) To find the pH at the equivalence point, first calculate the molarity of the NH4+ in the flask at this point.

\(\dfrac{0.054 \; mol \; NH_4{^+}}{0.140 \; L \; analyte \; solution}=0.375M \; NH_4{^+}\)

\(K_a=\dfrac{K_w}{K_b}=\dfrac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}}=5.56 \times 10^-10\)

\(5.56 \times 10^{-10}=\dfrac{x^2}{0.375+x}\)

\(2.09 \times 10^{-10}+(5.56 \times 10^{-10})x-x^2=0\)

We can use the quadratic equation to solve for x:

\(x=\dfrac{-5.56 \times 10^{-10} \pm \sqrt{(5.56 \times 10^{-10})^2-4(-1)(2.09 \times 10^{-10})}}{2(-1)}\) \(=\dfrac{-5.56 \times 10^{-10} \pm 2.89 \times 10^{-5}}{-2} = -1.45 \times 10^{-5}, \; 1.45 \times 10^{-5} \; M \; H_3O^+\)

\(pH=-log(1.45 \times 10^{-5})=4.84 \; pH\)

f) First, find the moles of HCl in 60 mL of HCl.

\(0.06 \; L \times \dfrac{mol\; HCl}{L \; HCl}=0.06 \; mol \; OH^-\)

Find the excess amount of HCl, or the amount added after neutralization has occurred.

0.054 moles of HCl reacted with the NH3 to neutralize it.

\(excess \; HCl=0.06-0.054=0.006 \; mol \; HCl\)

Now we need to find the molarity of HCl in the flask at this point. We started out with 90 mL of NH3 analyte in the flask, and added 60 mL. That gives a total of 150 mL, or 0.150 L of solution in the flask.

\(MOLARITY_{HCl \; in \; flask}=\dfrac{0.006 \; mol}{0.150 \; L \; solution}=0.04 \; M \; HCl\)

Because HCl dissociates into H3O+, equate [HCl] to [H3O+]. Now we have the information to determine pH.

\(pH=-log(0.04)=1.4 \; pH\)

The manufacture of soap requires a number of chemistry techniques. One necessary piece of information is the saponification number. This is the amount of base needed to hydrolyze a certain amount of fat to produce the free fatty acids that are an essential part of the final product. The fat is heated with a known amount of base (usually \(\ce{NaOH}\) or \(\ce{KOH}\)). After hydrolysis is complete, the leftover base is titrated to determine how much was needed to hydrolyze the fat sample.

At the equivalence point in a neutralization, the moles of acid are equal to the moles of base.

\[\text{moles acid} = \text{moles base}\nonumber \]

Recall that the molarity \(\left( \text{M} \right)\) of a solution is defined as the moles of the solute divided by the liters of solution \(\left( \text{L} \right)\). So the moles of solute are therefore equal to the molarity of a solution multiplied by the volume in liters.

\[\text{moles solute} = \text{M} \times \text{L}\nonumber \]

We can then set the moles of acid equal to the moles of base.

\[\text{M}_A \times \text{V}_A = \text{M}_B \times \text{V}_B\nonumber \]

\(\text{M}_A\) is the molarity of the acid, while \(\text{M}_B\) is the molarity of the base. \(\text{V}_A\) and \(\text{V}_B\) are the volumes of the acid and base, respectively.

Suppose that a titration is performed and \(20.70 \: \text{mL}\) of \(0.500 \: \text{M} \: \ce{NaOH}\) is required to reach the end point when titrated against \(15.00 \: \text{mL}\) of \(\ce{HCl}\) of unknown concentration. The above equation can be used to solve for the molarity of the acid.

\[\text{M}_A = \frac{\text{M}_B \times \text{V}_B}{\text{V}_A} = \frac{0.500 \: \text{M} \times 20.70 \: \text{mL}}{15.00 \: \text{mL}} = 0.690 \: \text{M}\nonumber \]

The higher molarity of the acid compared to the base in this case means that a smaller volume of the acid is required to reach the equivalence point.

The above equation works only for neutralizations in which there is a 1:1 ratio between the acid and the base. The example below demonstrates the technique to solve a titration problem for a titration of sulfuric acid with sodium hydroxide.

In a titration of sulfuric acid against sodium hydroxide, \(32.20 \: \text{mL}\) of \(0.250 \: \text{M} \: \ce{NaOH}\) is required to neutralize \(26.60 \: \text{mL}\) of \(\ce{H_2SO_4}\). Calculate the molarity of the sulfuric acid.

\[\ce{H_2SO_4} \left( aq \right) + 2 \ce{NaOH} \left( aq \right) \rightarrow \ce{Na_2SO_4} \left( aq \right) + 2 \ce{H_2O} \left( l \right)\nonumber \]

Solution

Step 1: List the known values and plan the problem.

• Molarity \(\ce{NaOH} = 0.250 \: \text{M}\)
• Volume \(\ce{NaOH} = 32.20 \: \text{mL}\)
• Volume \(\ce{H_2SO_4} = 26.60 \: \text{mL}\)

First determine the moles of \(\ce{NaOH}\) in the reaction. From the mole ratio, calculate the moles of \(\ce{H_2SO_4}\) that reacted. Finally, divide the moles of \(\ce{H_2SO_4}\) by its volume to get the molarity.

\[\begin{align*} &\text{mol} \: \ce{NaOH} = \text{M} \times \text{L} = 0.250 \: \text{M} \times 0.03220 \: \text{L} = 8.05 \times 10^{-3} \: \text{mol} \: \ce{NaOH} \\ &8.05 \times 10^{-3} \: \text{mol} \: \ce{NaOH} \times \frac{1 \: \text{mol} \: \ce{H_2SO_4}}{2 \: \text{mol} \: \ce{NaOH}} = 4.03 \times 10^{-3} \: \text{mol} \: \ce{H_2SO_4} \\ &\frac{4.03 \times 10^{-3} \: \text{mol} \: \ce{H_2SO_4}}{0.02660 \: \text{L}} = 0.151 \: \text{M} \: \ce{H_2SO_4} \end{align*}\nonumber \]

The volume of \(\ce{H_2SO_4}\) required is smaller than the volume of \(\ce{NaOH}\) because of the two hydrogen ions contributed by each molecule.

Summary

• The process of calculating concentration from titration data is described and illustrated.

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