Description Usage Arguments Value Author(s) Examples
This function calculates and plots the precipitation titration curve for an analyte and a titrant that form a precipitate with a 1:1 stoichiometry. The calculation uses a single master equation that finds the volume of titrant needed to achieve a fixed concentration of titrant, expressed as pTitrant, as outlined in R. de Levie's Principles of Quantitative Chemical Analysis (McGraw-Hill, 1997).
1 2 |
conc.analyte |
Molar concentration of the analyte; defaults to 0.025 M. |
conc.titrant |
Molar concentration of the titrant; defaults to 0.050 M. |
vol.analyte |
Initital volume, in mL, of the solution containing the analyte; defaults to 50.00 mL. |
pksp |
The pKsp value for the precipitate; defaults to 16.08, which is the pKsp for AgI. |
plot |
Logical; if TRUE, plots the titration curve. |
eqpt |
Logical; if TRUE, draws a vertical line at the titration curve's equivalence point. |
overlay |
Logical; if TRUE, adds the current titration curve to the existing titration curve. |
... |
Additional arguments to pass to |
A two-column data frame that contains the volume of titrant in the first column and the solution's pTitrant in the second column. Also produces a plot of the titration curve with options to display the equivalence point and to overlay titration curves.
David T. Harvey, DePauw University. harvey@depauw.edu
1 2 3 4 5 6 7 8 | ### Simple titration curve with equivalence point
ex14 = ppt_titrant(eqpt = TRUE)
head(ex14)
### Overlay titration curves using different pKsp values
ppt_titrant(pksp = 16, eqpt = TRUE)
ppt_titrant(pksp = 14, overlay = TRUE)
ppt_titrant(pksp = 12, overlay = TRUE)
|
volume p.titrant
1 49.47659 1.91
2 48.74305 1.92
3 48.03659 1.93
4 47.35591 1.94
5 46.69979 1.95
6 46.06710 1.96
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