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16-11 Lewis Acids and Bases -- Part 6
The pH rises slowly beyond the equivalence point.
The titration FIGURE 17-9 curve is essentially the same as Figure 17-8 if we plot pOH against the Titration curve for the strong acid.
We can make a set of statements that are similar to the titration of a strong base listed above.
For equal volumes of acid solutions of the same molarity, the volume of base required to reach the equivalence point is not dependent on the strength of the acid.
The equilibrium constant for the neutralization of reaction is the product of Ka CH COOH by NaOH.
CH3COO- + H2O Titration Data is done.
It will be very small.
To make a solution, NaOH to water.
Most of the OH will be consumed.
6.12 mol and 0.030 mol, respectively.
OH cannot have an equilibrium concentration of 0mol/L.
If 0 mol/L is theOH, this condition cannot be satisfied.
The fact that OH- is not completely consumed is a reminder that no reaction goes all the way to completion.
There are several options for adjusting the estimates.
Imagine if you will that the reaction above "backs up" a little bit to attain a true equilibrium state.
A different equilibrium calculation is another option.
The equilibrium calculation should be performed if the neutralization reaction goes to completion.
The neutralization reaction produces a solution that is, to a very good approximation, 0.070 M in CH COOH and 0.030 M in CH COO-.
The true equilibrium state can be determined by considering the ionization of CH COOH in the presence of an initial excess of NaCH COO.
We've used a line of reasoning before to simplify calculations.
0.10 mol NaOH to water to make a solution.
The calculation is based on the neutralization reaction.
The COO is going to water to maketions.
The neutralization reaction does not go all the way to completion.
Titrations between weak acids and strong bases are of interest.
The initial pH is the same as the pH for a solution of a weak acid or weak base.
The buffer region is the second, the third is the hydrolysis region and the fourth is beyond the equivalence point.
The total solution original acid + volume is 25.00 mL.
This information is entered into the following setup.
The Henderson-Hasselbalch equation can be used to calculate the pH of the acetic acid-sodium acetate solution.
When we added 12.50 mL of 0.
100 M NaOH, we added 1.25 MMol OH-.
At the equivalence point, neutralization is complete and 2.50mmol NaCH3COO has been produced in 50.00 mL of solution.
The amount of OH- added is 2.60 liters.
The 2.60 MMol OH 25.00 mL acid + 26.00 MMol base is 51.00 mL.
The excess strong base is what determines the solution's pH.
It was in the strong acid-strong base problem.
The result could have been predicted.
The conjugate base of a weak acid is CH 3COO.
A sample of 0.150 M HF solution is tested with 0.250 M NaOH.
Phenolphthalein is a good indicator, but methyl red is not.
Here are the main features of the curve for a weak acid with a strong base.
The initial pH is less acidic than a strong acid.
At the start of the titration, there is a sharp increase in the pH.
The pH changes slowly over a long section of the curve.
The point at which the pH 7 7 is equivalent.
The titration curve is the same as that of a strong acid with a strong base.
There is a steep portion of the curve at the equivalence point.
In a strong acid-strong base titration, there are more indicators to choose from.
The amount of strong base types depends on the portion of the curve being described.
A weak acid with a weak base is one type of titration that can't be done successfully.
The change in pH with volume of titrant is too gradual to find the equivalence point.
This stepwise neutraliza is the most striking evidence that a polyprotic acid ionizes in distinct steps.
We expect to see a separate tion for a polyprotic acid if there is an equal amount of acidic hydrogen.
When H and Ka differ, we expect to see three equiv cessive ionization constants.
All NaH2PO4 is converted to or more.
The Na2HPO4 is converted to Na3PO4.
The first two points are equal, and so on.
The third equivalence point is not realized in this titration.
Adding 0.100 M NaOH to water is not enough to reach the strongly hydrolyzed Na3PO4 solution at the third equivalence point.
As we will see in Section 17-5, Na3PO4(aq) is nearly as basic as the NaOH(aq) used in the titration.
There are a few details of this titration.
1 mol NaOH is required to reach the first equivalence point.
The solution is essentially NaH2PO4(aq).
To reach the first equivalence point, a volume of 0.
100 M NaOH is required.
The color of the orange indicator changes from red to orange when the pH at the equivalence point is within the range.
An additional mole of NaOH is needed to convert 1mol H 2PO4 to 1mol HPO 2 4.
The solu tion is basic because Kb 7 Ka for HPO 2 4 is the second equivalence point.
The indicator of this equivalence point is phenolphthalein, which is a light pink color.
All important points on the titration curve should be labeled.
We found that the first equivalence point should be in an acidic solution and the second in a mildly basic solution.
The third equivalence point could only be reached with a basic solution.
The third equivalence point is easy to calculate.
The equation x2 + 0.024x has a yield of 0.15 and a length of 0.015 M.
The approximation is valid.
The pH of 1.0 M Na2CO3 is calculated using data from Table 16.5.
The pH of 0.500 M Na2SO3 is calculated using data from Table 16.
This is due to both H Na3PO4
The general problem-solving method can be used here.
The following equations can be written.
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