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#' Precipitation Titration Curve: Mixture of Analytes
#'
#' This function calculates and plots the precipitation titration curve
#' for a mixture of two analytes using a titrant that form precipitates
#' with 1:1 stoichiometries. 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 \emph{Principles of Quantitative Chemical Analysis}
#' (McGraw-Hill, 1997).
#'
#' @param conc.analyte1 Molar concentration of the first analyte;
#' defaults to 0.050 M.
#'
#' @param conc.analyte2 Molar concentration of the second analyte;
#' defaults to 0.050 M.
#'
#' @param conc.titrant Molar concentration of the titrant;
#' defaults to 0.050 M.
#'
#' @param vol.analyte The initial olume, in mL, of the solution
#' containing the analyte; defaults to 25.00 mL.
#'
#' @param pksp1 The pKsp value for the first analyte's precipitate;
#' defaults to 16.08, which is the pKsp for AgI.
#'
#' @param pksp2 The pKsp value for the second analyte's precipitate;
#' defaults to 11.97, which is the pKsp for AgSCN.
#'
#' @param plot Logical; if TRUE, plots the titration curve.
#'
#' @param eqpt Logical; if TRUE, draws a vertical line at the titration
#' curve's equivalence point.
#'
#' @param overlay Logical; if TRUE, adds the current titration curve
#' to the existing titration curve.
#'
#' @param \dots Additional arguments to pass to \code{plot()} function.
#'
#' @return 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.
#'
#' @author David T. Harvey, DePauw University. \email{harvey@@depauw.edu}
#'
#' @export
#'
#' @importFrom graphics plot lines
#'
#' @examples
#' ### Simple titration curve with equivalence points
#' ex15 = ppt_mixture(eqpt = TRUE)
#' head(ex15)
#'
#' ### Overlay mixture titration curves using different pKsp values
#' ppt_mixture(pksp1 = 16, pksp2 = 12, eqpt = TRUE)
#' ppt_mixture(pksp1 = 14, pksp2 = 10, overlay = TRUE)
ppt_mixture = function(conc.analyte1 = 0.05, conc.analyte2 = 0.05,
vol.analyte = 25, conc.titrant = 0.05,
pksp1 = 16.08, pksp2 = 11.97, plot = TRUE,
eqpt = FALSE, overlay = FALSE, ...) {
veq1 = conc.analyte1 * vol.analyte/conc.titrant
veq2 = conc.analyte2 * vol.analyte/conc.titrant
veq = veq1 + veq2
ksp1 = 10^-pksp1
ksp2 = 10^-pksp2
y.lim = max(c(pksp1, pksp2))
p.titrant = seq(1, y.lim, 0.01)
titrant = 10^-p.titrant
analyte1 = ksp1/titrant
analyte2 = ksp2/titrant
volume1 = vol.analyte * (conc.analyte1 - analyte1 + titrant)/
(conc.titrant + analyte1 - titrant)
volume2 = vol.analyte *
(conc.analyte1 + conc.analyte2 - analyte1 - analyte2 + titrant)/
(conc.titrant + analyte1 + analyte2 - titrant)
df = data.frame(volume1, volume2)
volume = apply(df, 1, max)
df = data.frame(volume, p.titrant)
df = df[df$volume > 0 & df$volume < 2 * veq, ]
rownames(df) = 1:nrow(df)
if (plot == TRUE) {
if (overlay == FALSE) {
plot(df$volume, df$p.titrant, type = "l", lwd = 2,
xlim = c(0, 1.5 * veq), ylim = c(0, y.lim),
xlab = "volume of titrant (ml)", ylab = "pTitrant",
xaxs = "i", yaxs = "i", ...)
} else {
lines(df$volume, df$p.titrant, type = "l", lwd = 2, ...)
}
if (eqpt == TRUE) {
x1 = c(veq1, veq1)
x2 = c(veq, veq)
y = c(-1, y.lim + 1)
lines(x1, y, type = "l", lty = 2, col = "red")
lines(x2, y, type = "l", lty = 2, col = "red")
}
}
invisible(df)
}
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