Titrations in Chemistry : Titrations in Chemistry To date we have worked on calculating concentrations of the components of a system as a function of concentration of one component.
With a few simple formulas we can extend all these ideas in order to treat titrations. Read Chapter 7-4 to 7-7.
Slide2 : Consider the addition of AgNO3 to a solution of NaI. Goal is to calculate the concentration pKAg=-log[Ag+] as a function of the volume Vt of AgNO3 added. Vt pKAg Goal is to calculate [Ag+] as a function of Vt.
Conservation of Moles : Conservation of Moles Initial moles of I- = Final moles of I- Moles solid AgI Initial moles of Ag+ = Final moles of Ag+ Moles solid AgI Vi=initial volume of NaI solution
Vt=volume of AgNO3 added
Slide4 : Equating moles of solid AgI Rearrange terms so all terms that depend on Vt are on the right and all terms that depend on Vi are on the left Finally solve for Vt.
Solve for Vt : Solve for Vt We have two unknowns and two equations. We now turn to numerical calc. to make our plots. Insert A. Using the second equation write Vt as function of [Ag+].
Titration Plot for AgI : Titration Plot for AgI Before discussing this plot, I want to return to the equations we used to derive it, to show you another approach.
Slide7 : We saw before, that equating moles of solid AgI leads to Dividing both sides by Vi+Vt gives Recognize that the circled terms can be rewritten to give This is just the charge balance equation. This Eq. can be derived
from the CB Eq.
Titration Plot for AgI : Titration Plot for AgI By understanding this plot, we will see that there is an easy
way to obtain analytical solutions. In the green region there is excess I-. In the white region there is excess Ag+.
Green Region Calculation : Green Region Calculation Add 21.00 mL of .0500AgNO3 to 50.00 mL of .0500M NaI. Convert to concentrations and solve Ksp problem using ice tables. Before .01479 .03521 ns 0 .03521-.01479 solid After x .02042+x solid Ksp=8.3x10-17
=x(.02042) pKAg = 14.4
Titration Plot for AgI : Titration Plot for AgI By understanding this plot, we will see that there is an easy
way to obtain analytical solutions. In the green region there is excess I-. In the white region there is excess Ag+.
White Region Calculation : White Region Calculation Add 51.00 mL of .05000AgNO3 to 50.00 mL of .05000M NaI. Convert to concentrations and solve Ksp problem using ice tables. Before .025248 .024752 solid .025248-.024752 0 solid After .000496+x x solid Ksp=8.3x10-17
=x(.000496) pKAg = 3.30
Conclusions : We have seen that titrations involving precipitation reactions are easy to solve if we are
1)careful about the region we are in and
2) set up the ICE table accordingly.
These problems can also be solved with numerically by combining CB and Ksp. Here we don’t have to worry about which region we are in.
Now lets move on to more complex problems. Conclusions
A More Complex Titration Problem : A More Complex Titration Problem Add Vt mL of .02500M AgNO3 to 50.00 mL of .05000M NaI + 50.00 mL of .05000M NaCl Begin with CB Eq. Instead of deriving all the equations again, lets see if we can get the solution by inspection. Lets review what we did earlier. See Chapter 7-5.
Slide14 : We saw before that equating moles of solid AgI leads to Dividing both sides by Vi+Vt gives Recognize that the circled terms can be rewritten to give This is just the charge balance equation. This Eq. can be derived
from the CB Eq.
Slide15 : From the CB equation we had Rearranging terms led to By inspection the
answer to the new
problem is Final step is to use Ksp so that everything depends on [Ag+].
Numerical Titration Curve : Numerical Titration Curve 1 2 3
Approximate Approach : Add Vt mL of .02500M AgNO3 to 50.00 mL of .05000M NaI + 50.00 mL of .05000M NaCl Solve for the initial concentrations if Vt=100.00mL. Approximate Approach [Ag+]i=100/200*.0250=.0125M
[I-]i = [Cl-]i =100/200*.0250=.0125M
Slide18 : Approximate Approach (Continued) .0125 .0125 ns .0125 .0125 ns 0 0 solid 0 .0125 ns x x solid x .0125 ns Solve for x using KspAgI =8.3x10-17=x2 yields x = 9.1x10-9 Note x(.0125)Approximate Approach (Continued) : Add Vt mL of .02500M AgNO3 to 50.00 mL of .05000M NaI + 50.00 mL of .05000M NaCl Solve for the concentrations when Vt=150.00mL. Approximate Approach (Continued) [Ag+]i=150/250*.0250=.015M
[I-]i = [Cl-]i =100/250*.0250=.010M
Slide20 : Approximate Approach (Continued) .015 .010 ns .015 .010 ns 0 0 solid 0 .005 ns 0 0 solid y .005+y ns Using KspAgI =8.3x10-17=3.6x10-8z yields z = 2.3x10-9 Solve for y using KspAgCl =1.8x10-10=.005y yields y = 3.6x10-8
Approximate Approach (Continued) : Add Vt mL of .02500M AgNO3 to 50.00 mL of .05000M NaI + 50.00 mL of .05000M NaCl Solve for the initial concentrations if Vt=100.10mL. Approximate Approach (Continued) [Ag+]i=100.1/200.1*.0250=.012506M
[I-]i = [Cl-]i =100/200.1*.0250=.012493M
Approximate Approach (Continued) : .012506 .012494 ns .012506 .012494 ns 0 0 solid 0 .01248 ns 0 0 solid y .01248+y ns Solve KspAgCl =1.8x10-10 =.01248y yields y = 1.4x10-8 Approximate Approach (Continued)
Complex Titration Problems : Complex Titration Problems In order to solve these problems, you should:
1) make a rough sketch of the titration plot,
2) identify the reasons, and finally
3) calculate the concentrations in each region. You should review problems 7-26 to 7-28 to see if you can solve them.