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A titration is a controlled chemical reaction between two different solutions.
In this reaction a buret is used to administer one solution to another. The solution administered from the buret is called the titrant. The solution that the titrant is added to is called the analyte. In a titration of a Weak Acid with a Strong Base the titrant is a strong base and the analyte is a weak acid. In order to fully understand this type of titration the reaction, titration curve, and type of titration problems will be introduced.
Figure 1: Titrations involve the addition of the titrant from the burret to the analyte. Figure is used with the permission of J.A. Freyre under the Creative Commons Attributions-Share Alike 2.5 Generic
The titration of a weak acid with a strong base involves the direct transfer of protons from the weak acid to the hydoxide ion. The reaction of the weak acid, oxalic acid, with a strong base, NaOH, can be seen below. In the reaction the acid and base react in a one to one ratio.
\(H_2C_2O_4 + OH^- \longrightarrow H_2O + HC_2O_4^-\)
The titration curve is a graph of the volume of titrant, or in our case the volume of strong base, plotted against the pH. There are several characteristics that are seen in all titration curves of a weak acid with a strong base. These characteristics are stated below.
The image of a titration curve of a weak acid with a strong base is seen below. All of the characteristics described above can be seen within it.
Figure 2: The titration of a weak acid with strong base. Figure is used with the permission of J.A. Freyre under the Creative Commons Attributions-Share Alike 2.5 Generic
When solving a titration problem with a weak acid and a strong base there are certain values that you want to attain. These include the initial pH, the pH after adding a small amount of base, the pH at the half-neutralization, the pH at the equivalence point, and finally the pH after adding excess base. This data will give sufficient information about the titration. Below is an example of this process.
Find the pH at each of the following points in the titration of 25 mL of 0.3 M HF with 0.3 M NaOH. The k_{a} value is 6.6*10^{-4}
Example 1: The initial pH | ||||||||||||||||||||
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Finding the initial pH. SOLUTION Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.
Writing the information from the ICE Table in Equation form yields \(6.6*10^{-4} = \dfrac{x^{2}}{.3-x}\) Manipulating the equation to get everything on one side yields \(0 = x^{2} + 6.6*10^{-4}x - 1.98*10^{-4}\) Now this information is plugged into the quadratic formula to give \(x = \dfrac{-6.6*10^{-4} \pm \sqrt{(6.6*10^{-4})^2 - 4(1)(-1.98*10^{-4})}}{2}\) The quadratic formula yields that x=.013745 and x=-.014405 However we can rule out x=-.014405 because there cannot be negative concentrations Therefore to get the pH we plug the concentration of H_{3}O^{+} into the equation pH=-log(.013745) and get pH=1.86 |
Example 2: After adding 10 mL of 0.3 M NaOH | |||||||||||||||||||||||||
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Find the pH after the addition of 10 mL of 0.3 M NaOH. SOLUTION
To calculate the pH with this addition of base we must use an ICE Table
However, this only gives us the millimoles. To get the concentration we must divide by the total volume. The total volume is the 25 mL original soultion of HF plus the 10 mL of NaOH that was added. Therefore, the total volume is \(25 mL + 10 mL = 35 mL\)
Since an acid and its conjugant base are in equilibrium we can attempt to use the Henderson-hasselbalch equation. However, for this to work the reaction must follow certain rules. The ratio of the conjugant base and weak acid must be between 0.10 and 10. Also, both the ratio of the conjugant base and k_{a} value and the ratio of the acid and k_{a} value must exceed 100. In this problem the Henderson-hasselbalch equation can be applied because the ratio of F- to HF is \(\frac{.0857}{.1287} = 0.666\) . This is between 0.10 and 10. The ratio of HF to k_{a} is \(\frac{.1287M}{6.6*10^{-4}} = 195\) and the ratio of F^{-} to k_{a} is \(\frac{.0857M}{6.6*10^{-4}} =130\). These both exceed one hundred. Therefore, we continue by using the Henderson-hasselbalch equation. \(pH=pk_{a} + log\dfrac{[A^{-}]}{[HA]}\) \(pH=-log(6.6*10^{-4}) + log\dfrac{.0857}{.1287}\) pH=3.00 |
Example 3: After adding 12.50 mL of 0.3 M NaOH | |||||||||||||||||||||||||
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Find the pH after adding 12.50 mL of 0.3 M NaOH. SOLUTION The millimoles of OH^{-} added in 12.50 mL: \(12.50 mL * \dfrac{.3 mmol OH^{-}}{mL} =3.75 mmol OH^{-}\) Once again we must use an ICE Table
To find the concentrations we must divide by the total volume. This is the initial volume of HF, 25 mL, and the addition of NaOH, 12.50 mL. Therefore the total volume is 25 mL + 12.50 mL = 37.50 mL
We have found the Half-neutralization point. We know this because the total amount of acid to be neutralized, 7.50mmol, has been reduced to half of its value, 3.75 mmol. At the half-neutralization point we can simplify the henderson-hasselbalch equation and use it. Since the amount of conjugant base and acid are equal, their ratio is one. We know that \(log(1) =0\) and therefore the ratio of conjugant base to acid will be zero as well. The equation at the half-neutralization point will be \(pH=pk_{a} +log(1)\) which equals \(pH=pk_{a}\) \(pH=-log(6.6*10^{-4})\) pH=3.18 |
Example 4: After adding 25 mL of 0.3 M NaOH | |||||||||||||||||||||||||||||||||||||||||||||
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Find the pH after the addition of 25 mL of NaOH. SOLUTION
This is the equivalence point of the titration. We know this because the acid and base are both neutralized and neither is in excess. To find the concentrations we must divide by the total volume. This is the initial volume of HF, 25 mL, and the addition of NaOH, 25 mL. Therefore the total volume is 25 mL + 25 mL = 50 mL Concentration of F^{-}:\(\dfrac{7.5 mmol F^{-}}{50 mL}=.15M\) However, to get the pH at this point we must realize that F^{-} will hydrolyze. An ICE table for this reaction must be constructed
In this reaction the F^{-} acts as a base. Therefore we must obtain the k_{b} value instead of the k_{a} value. \(k_{b}= \dfrac{k_{w}}{k_{a}}\) \(k_{b} = \dfrac{1.0*10^{-14}}{6.6*10^{-4}}\) \(k_{b}=1.515*10^{-11}\) Now that we have the k_{b} value, we can write the ICE table in equation the equation form \(1.515*10^{-11} \dfrac{x^{2}}{.15-x}\) Manipulating the equation to get everything on one side yields \(0= x^{2} + 1.515*10^{-11}x -2.2727*10^{-12}\) Now this information is plugged into the quadratic formula to give \(x = \dfrac{-1.515*10^{-11} \pm \sqrt{(-1.515*10^{-11})^2 - 4(1)(-2.2727*10^{-12})}}{2}\) The quadratic formula yields x=1.5075*10^{-6} and -1.5075*10^{-6} . However the negative value can be ruled out because concentrations cannot be zero. Therefore to get the pOH we plug the concentration of OH^{-} into the equation pH=-log(1.5075*10^{-6}) and get pOH=5.82. To get the pH we minus the pOH from 14. pH=14 - 5.82 pH= 8.18 |
Example 5: After adding 26 mL of 0.3 M NaOH |
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Find the pH after the addition of 26 mL of NaOH. SOLUTION The millimoles of OH^{-} added in the 26 mL: \(26 mL * \dfrac{.3 mmol OH^{-1}}{1 mL} = 7.8 mmol OH^{-}\) This amount is greater then the moles of acid that is present. The 7.8 mmol OH^{-} neutralizes the 7.50 mmol HCl. To find how much OH^{-} will be in excess we subtract the amount of acid and hydroxide. mmoles of hydroxide in excess: 7.8 mmol - 7.50 mmol= .3 mmol OH^{-} To find the concentration of the OH^{-} we must divide by the total volume. This is the initial volume of HF, 25 mL, and the addition of NaOH, 26 mL. Therefore the total volume is 25 mL + 26 mL = 51 mL The concentration of OH^{-} is \(\dfrac{.3 mmol OH^{-}}{51 mL}=.00588M\) pOH=-log(0.00588)=2.23 pH=14-2.23 pH=11.77 |
Example 6 |
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When does the equivalence point of 15 mL of 0.15 M CH_{3}COOH titrated with 0.1 M NaOH occur? SOLUTION The equivalence point occurs when equal moles of acid react with equal moles of base. The mmol CH_{3}COOH: \(15 mL CH_{3}COOH * \dfrac{.15 mmol CH_{3}COOH}{1 mL} =2.25 mmol CH_{3}COOH\) We must find the amount of of mL of NaOH to give us the same mmols as CH_{3}COOH \(2.25 mmol CH_{3}COOH = 0.1M NaOH* XmL NaOH\) X=22.5 mL Therefore the equivalence point is after the addition of 22.5 mL of NaOH |
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