Abstract

In this experiment we used an autotitrator that added a known base to an unknown acid. By titrating a base and knowing the volume of and the concentration of the base we can calculate the molecular weight of the unknown acid. The dissociation constant can also be calculated of the unknown acid. With the molecular weight and dissociation constant we can determine what the unknown acid is.

Introduction

Titration is a technique that is used to determine the concentration of an acid by neutralizing it with a base. In our laboratory class two experiments were conducted where we titrated a base with an unknown acid. In the first experiment we titrated a base with a monoprotic acid. A monoprotic acid is an acid that donates one hydrogen ion or proton per molecule when it dissociates in an aqueous solution. As a base was slowly added to the to the acid the acid starts to give up its protons becoming less acidic and raising the pH value of the solution. The pKa value is the value when half of the acid molecules are deprotonated. The endpoint or equivalence point is the value when all of its protons are deprotonated. The second experiment that was done a diprotic acid was used. A diprotic acid is an acid that can donate two hydrogen ions or protons per molecule in an aqueous solution. An auto titrator was used in this experiment and software collected the volume and pH data and that data was transferred to excel where a monoprotic graph and a diprotic graph was generated.

Purpose

The purpose of this paper is to take the data that was obtained from the two experiments and use Posit to graph the monoprotic and diprotic curves. Posit was also used to graph the binding curve to determine the pKa values.

Monoprotic titration analysis

A monoprotic titration curve shows a single equivalance point. The graph is represented by a pH in the y-axis and the volume of the base NaOH in the x-axis.

Explain how data was transformed to binding curves, using equations, and graph binding curve.

The two equations below was used to generate the fraction bound and best fit line. CB is the base NaOH that was titrated which is 0.10 M. The volume of acid added is Vadd. The initial volume is 25 mL and was represented by Vini. Vend is the ending volume and it was 19.5mL. A graph was generated that showed the binding curve. The pH is in the x-axis and the fraction bound is in the y-axis.

\(f = 1−(CB×Vadd+[H+]×(Vini+Vadd)/{CB×Vend}\)

\(F= \frac {H}{H+CB}\)

The value of Ka

The value of Ka is determined by the fit and is 3.130 x 10^-5.

Compare the traditional titration analysis to a binding curve. What are pros and cons.

The pros of a binding curve it uses mathematical analysis to estimate the Ka value to a high degree of accuracy. The pros of using titration is that it allows for precise and accurate determination of the Ka value.

Data <- read.csv("Titration analysis.csv")

Volume <- Data$ml

pH <-Data$pH

plot(Volume,pH,main = "Volume of NaOH vs. pH",xlab = "Volume of NaOH (mL)",ylab = "pH")

#Proton Concentration 

H <- 10^(-pH) 

#Base Concentration

BC <- 0.01

#Initial Acid Volume

Vini <- 25

#Estimated Volume of Base Needed to reach Endpoint

Vend <- 19.5

#Amount of Volume Added

Vadd <- Volume

#Fraction Bound

FB <- 1 - ((BC * Vadd) + H  * (Vini + Vadd)) / (BC * Vend)

FB
##  [1]  0.9049602255  0.9135668636  0.8732072510  0.8281600242  0.7810563973
##  [6]  0.7292979530  0.6810531696  0.6319450238  0.5822708591  0.5322034124
## [11]  0.4815685755  0.4315496526  0.3809156795  0.3303359038  0.2796133029
## [16]  0.2288680414  0.1780492114  0.1527621933  0.1272518609  0.1017617702
## [21]  0.0763101145  0.0508520650  0.0253746969 -0.0001268602 -0.0256631147
## [26] -0.0512821337 -0.0769230925 -0.1025641093 -0.1282051322 -0.1538461567
## [31] -0.1794871818 -0.2051282069 -0.2307692324 -0.2820512834 -0.3333333345
## [36] -0.3846153856 -0.4358974368 -0.4871794880 -0.5384615392 -0.5897435905
## [41] -0.6410256417
plot(pH,FB, main = "Binding Curve",xlab = "pH",ylab = "Fraction Bound",pch=20,col="red")

library(nls2)
## Loading required package: proto
fit <- nls2(FB ~ H/(KD+H), start=c(KD=0.0001))

summary(fit)
## 
## Formula: FB ~ H/(KD + H)
## 
## Parameters:
##     Estimate Std. Error t value Pr(>|t|)    
## KD 3.130e-05  8.332e-06   3.757 0.000549 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.2241 on 40 degrees of freedom
## 
## Number of iterations to convergence: 4 
## Achieved convergence tolerance: 1.832e-06
lines(pH,predict(fit),col="blue")

Transforming Diprotic Data. Show and explain the equation.

The equation below is the equation for the fraction bound for a diprotic acid and the fit line formula. The concentration of the base is represent by tbase and was 0.10. The inital volume of the acid is 25 mL. The volume at the end of the second end point is 9.3025.

\(Fb <- (2-(((Volume*tbase)+((H)*(Vini+Volume)))/(Vend*tbase)))\)

Fit line formula is

\(F= \frac {H}{H+CB}\)

Use NLS2 to determine Ka1 and Ka2

The Ka1 and Ka2 values from NlS2 is. Ka1 = 6.17x10^-7 Ka2 = 9.945x10^-4

Compare the Kas from two different methods

The Ka values from the experiment that were obtained from a graph of the the titration curve vs the volume of base added are. Which are different values from the Ka1 and Ka2 from NLS2.

Ka1= 10^(-3.9)=0.000125 Ka2=10^(-6.79)=1.62 x 10^-7

What further thoughts do you have for future studies.

Using R-Scirpt to analyze titration curves and graph its binding curves is a valuable tool to have for future experiments.

Diprotic

#Diprotic titration


Data <- read.csv("Diprotic.csv")

Volume <- Data$Volume

pH <- Data$pH

plot(Volume, pH, main = "Diprotic Acid Titration Curve", xlab = "Volume of NaOH (mL)", ylab = "pH",pch = 20, col = "blue")

H <- 10^-(pH)    ## H+ from pH

Vini <- 25           ## Initial Volume

Vend <- 9.3025        ## Volume at endpoint

tbase <- 0.10        ## Base

Fb <- (2-(((Volume*tbase)+((H)*(Vini+Volume)))/(Vend*tbase)))

library(nls2)

fit <- nls2(Fb~(H/KD1+2*H^2/(KD1*KD2))/(1+H/KD1+H^2/(KD1*KD2))
            ,start = c(KD1=0.001,KD2=0.000001))

summary(fit)
## 
## Formula: Fb ~ (H/KD1 + 2 * H^2/(KD1 * KD2))/(1 + H/KD1 + H^2/(KD1 * KD2))
## 
## Parameters:
##      Estimate Std. Error t value Pr(>|t|)    
## KD1 6.170e-07  6.478e-08   9.524 1.57e-15 ***
## KD2 9.945e-04  8.322e-05  11.950  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.1079 on 96 degrees of freedom
## 
## Number of iterations to convergence: 17 
## Achieved convergence tolerance: 4.783e-06
plot(pH,Fb)

lines(pH,Fb,col="purple")