Abstract

The purpose of this experiment is to learn how to titrate a diprotic acid using an autotitrator and using the data identify the acid. The autotitrator features titrator connections, high-wear components, an electrode holder, pH electrode technology, a bottle holder, color coded tubing and connectors, and a stirrer probe (Reference #2). An autotitrator is better than manual titration because it’s faster compared to manual titration, eliminates the handling of corrosive titrants, increases the reproducibility of an experiment, and increases the accuracy of the volume of titrant added (Reference #1) .Diprotic acids are acids that contain two hydrogen atoms that can be easily deprotonated. This means that there are two equivalence points.“At the first equivalence point, all H+ ions from the first dissociation have reacted with NaOH base” and at the second equivalence point there is no longer any H+ left so twice as many H+ from the first equivalence point have been neutralized. The hypothesis for this experiment was the identity of the diprotic acid will correspond to the acid that shares the same Ka1 and Ka2. To test this the Ka had to be found using an autotitrator and then the data was analyzed by using the first derivative test to find both equivalence points then going to the half equivalence of each and finding both the Ka from there. The data collected and analyzed shows that the Ka1= 6.457e-03 and Ka2=1.38e-05. This matches with the acid being Malonic acid because the Ka of Malonic acid are in the same range. Now we used R to compare how close our data is and which method is better.

Introduction

Last week we performed a potentiometric titration with a monoprotic acid using a pH meter and manual titration. We used that data to create a binding curve and calculate the Ka and compare it to the Ka we calculated manually. We know from the Monoprotic that the equivalence point is the point where the concentration of H⁺ is equal to the concentration of the OH⁻ and the half equivalence point is the point at which the concentration of the acid is equal to the concentration of the conjugate base because it is halfway neutralized.The Henderson-Hassleback equation is pH=pKa +log(base/acid) or in our case pH=pKa +log([CH₃COO⁻]/[CH₃COOH]). Since we understand that at the half equivalence point our conjugate base concentration is equal to our acid concentration that would mean log(1) which is equal to 0 so the pH=pKa at the half equivalence point.This also applies to diprotic acids. Diprotic acids are acids that contain two hydrogen atoms that can be easily deprotonated. The equation for diprotic acids are as follows

The 1st succession is the following equation: \(H2A + H2O <-> H3O+ + HA-\).
The second succession is the following equation: \(HA- + H2O <-> H3O+ + A2-\).

It has two equivalence points which is a result of the “successive dissociation” of diprotic acids as shown with the two equations above therefore creating two equivalence points. “At the first equivalence point, all H+ ions from the first dissociation have reacted with NaOH base” and at the second equivalence point there is no longer any H+ left so twice as many H+ from the first equivalence point have been neutralized.The Lab Report can be improved through more accurate calculations since the titrations aren’t exactly at the 1/2 of the pKa so its more of a close estimate. For more sophisticated data using a best fit line might be more accurate for calculations.

Titration Graph:


``` r
mydata <- read.csv('data.csv')

VolVect <- mydata$A
VolVect
##  [1]  0.000  0.005  0.010  0.023  0.054  0.132  0.328  0.528  0.728  0.928
## [11]  1.128  1.328  1.528  1.728  1.928  2.128  2.328  2.528  2.728  2.928
## [21]  3.129  3.329  3.529  3.729  3.929  4.129  4.329  4.529  4.729  4.929
## [31]  5.129  5.329  5.529  5.729  5.929  6.129  6.329  6.529  6.639  6.733
## [41]  6.807  6.885  6.977  7.078  7.163  7.336  7.536  7.736  7.936  8.136
## [51]  8.337  8.537  8.737  8.937  9.137  9.337  9.537  9.737  9.937 10.137
## [61] 10.337 10.537 10.737 10.937 11.137 11.337 11.537 11.737 11.937 12.137
## [71] 12.337 12.537 12.737 12.937 13.137 13.337 13.537 13.737 13.937 14.137
## [81] 14.337 14.537
pHVect <- mydata$pH
pHVect
##  [1] 1.84 1.84 1.85 1.85 1.85 1.85 1.86 1.88 1.89 1.91 1.93 1.94 1.96 1.99 2.02
## [16] 2.04 2.06 2.08 2.11 2.14 2.18 2.19 2.24 2.26 2.30 2.33 2.37 2.42 2.47 2.54
## [31] 2.60 2.64 2.70 2.82 2.91 3.04 3.23 3.48 3.65 3.84 4.00 4.16 4.34 4.55 4.67
## [46] 4.86 5.05 5.18 5.28 5.39 5.48 5.57 5.63 5.69 5.75 5.81 5.82 5.92 5.96 5.97
## [61] 6.07 6.10 6.15 6.19 6.23 6.28 6.33 6.38 6.43 6.44 6.53 6.58 6.63 6.71 6.78
## [76] 6.78 6.90 7.00 7.01 7.18 7.20 7.49
plot(VolVect,pHVect,xlab = 'Volume (ml)',ylab = 'pH',main = 'Titration Curve')

The graph above is our data for the diprotic titration graphed. As you can see there are two equivalent points.

Binding Curve:

We understand that as the reaction proceeds the H is being depronated so the concentration of H+ is increasing and what is left is the conjugate base which is also increasing as you continue removing the H from that conjugate base. With this we can calculate the fraction bound as the reaction proceeds (the volume is added). We can get the concentration of H+ by doing pH=-log([H+]), we can manipulate this equation to [H+]=10^-pH to get our H vector.

H <- 10^(-pHVect)
H
##  [1] 1.445440e-02 1.445440e-02 1.412538e-02 1.412538e-02 1.412538e-02
##  [6] 1.412538e-02 1.380384e-02 1.318257e-02 1.288250e-02 1.230269e-02
## [11] 1.174898e-02 1.148154e-02 1.096478e-02 1.023293e-02 9.549926e-03
## [16] 9.120108e-03 8.709636e-03 8.317638e-03 7.762471e-03 7.244360e-03
## [21] 6.606934e-03 6.456542e-03 5.754399e-03 5.495409e-03 5.011872e-03
## [26] 4.677351e-03 4.265795e-03 3.801894e-03 3.388442e-03 2.884032e-03
## [31] 2.511886e-03 2.290868e-03 1.995262e-03 1.513561e-03 1.230269e-03
## [36] 9.120108e-04 5.888437e-04 3.311311e-04 2.238721e-04 1.445440e-04
## [41] 1.000000e-04 6.918310e-05 4.570882e-05 2.818383e-05 2.137962e-05
## [46] 1.380384e-05 8.912509e-06 6.606934e-06 5.248075e-06 4.073803e-06
## [51] 3.311311e-06 2.691535e-06 2.344229e-06 2.041738e-06 1.778279e-06
## [56] 1.548817e-06 1.513561e-06 1.202264e-06 1.096478e-06 1.071519e-06
## [61] 8.511380e-07 7.943282e-07 7.079458e-07 6.456542e-07 5.888437e-07
## [66] 5.248075e-07 4.677351e-07 4.168694e-07 3.715352e-07 3.630781e-07
## [71] 2.951209e-07 2.630268e-07 2.344229e-07 1.949845e-07 1.659587e-07
## [76] 1.659587e-07 1.258925e-07 1.000000e-07 9.772372e-08 6.606934e-08
## [81] 6.309573e-08 3.235937e-08

To create the fraction bound we use the equation below.

Vi <- 25
Vi
## [1] 25
Vend <- 7.2
Vend
## [1] 7.2
Cbase <- 0.1
Cbase
## [1] 0.1
F<- (2- ((VolVect*Cbase)+ (H) *(Vi+VolVect) )/(Vend*Cbase))
F
##  [1]  1.498111191  1.497316369  1.507950500  1.505889903  1.500976172
##  [6]  1.488612589  1.468855935  1.459271416  1.438554382  1.428077657
## [11]  1.416976065  1.395714048  1.383786478  1.380130901  1.375054995
## [16]  1.360819027  1.346087597  1.330877874  1.322169722  1.312332674
## [21]  1.307296583  1.283600852  1.281851028  1.262808892  1.252932702
## [26]  1.237296432  1.224984018  1.215047046  1.203284751  1.195533085
## [31]  1.182526908  1.163361493  1.147481440  1.139708023  1.123679191
## [36]  1.109319465  1.095350163  1.078694121  1.068079042  1.058490536
## [41]  1.050165694  1.040686246  1.028942179  1.015688777  1.004183843
## [46]  0.980491165  0.952930587  0.925255160  0.897537708  0.869812515
## [51]  0.841930015  0.814180186  0.786417934  0.758653763  0.730887910
## [56]  0.703120581  0.675344064  0.647580885  0.619807906  0.592031042
## [61]  0.564263782  0.536488572  0.508714861  0.480939996  0.453164890
## [66]  0.425390181  0.397615153  0.369839841  0.342064273  0.314286828
## [71]  0.286512474  0.258736287  0.230959936  0.203184171  0.175407876
## [76]  0.147630052  0.119854373  0.092077953  0.064300271  0.036524186
## [81]  0.008746553 -0.019029555
tf <- F

The VolVect represents the volume added, Cbase is the concentration of the base, H is the H+ concentration, Vi is the initial volume and Vend is the volume at the equivalence point. The 2 represents the total amount of H we can depronate since there are 2H. The VolVectCbase is the mols of OH- (Lmols/L=mols) which is being added by the concentration of H+ Vi+VolVect which is the mols of H+ over VendCbase which is the total mols of the OH- at the equivalence point so its also the total mols of H+. This gives us the fraction bound of H+.

The graph ends up looking something like the following:

F<- (2- ((VolVect*Cbase)+ (H) *(Vi+VolVect) )/(Vend*Cbase))
F
##  [1]  1.498111191  1.497316369  1.507950500  1.505889903  1.500976172
##  [6]  1.488612589  1.468855935  1.459271416  1.438554382  1.428077657
## [11]  1.416976065  1.395714048  1.383786478  1.380130901  1.375054995
## [16]  1.360819027  1.346087597  1.330877874  1.322169722  1.312332674
## [21]  1.307296583  1.283600852  1.281851028  1.262808892  1.252932702
## [26]  1.237296432  1.224984018  1.215047046  1.203284751  1.195533085
## [31]  1.182526908  1.163361493  1.147481440  1.139708023  1.123679191
## [36]  1.109319465  1.095350163  1.078694121  1.068079042  1.058490536
## [41]  1.050165694  1.040686246  1.028942179  1.015688777  1.004183843
## [46]  0.980491165  0.952930587  0.925255160  0.897537708  0.869812515
## [51]  0.841930015  0.814180186  0.786417934  0.758653763  0.730887910
## [56]  0.703120581  0.675344064  0.647580885  0.619807906  0.592031042
## [61]  0.564263782  0.536488572  0.508714861  0.480939996  0.453164890
## [66]  0.425390181  0.397615153  0.369839841  0.342064273  0.314286828
## [71]  0.286512474  0.258736287  0.230959936  0.203184171  0.175407876
## [76]  0.147630052  0.119854373  0.092077953  0.064300271  0.036524186
## [81]  0.008746553 -0.019029555
tf <- F
plot(pHVect,F,xlab='pH',ylab='Fraction Bound',main='Binding Curve')

We added a fit line and then we followed this up by doing the summary of the fit to find our Ka values. Using the following theoretical equation:

\((H/Ka1 + 2*H^2/(Ka1*Ka2))/(1+H/Ka1 + H^2(Ka1*Ka2))\)

plot(pHVect,F,xlab='pH',ylab='Fraction Bound',main='Binding Curve')

library(nls2)
## Loading required package: proto
tryfit <- nls2(tf ~ (H/Ka1 + 2*H^2/(Ka1*Ka2))/(1+H/Ka1 + H^2/(Ka1*Ka2)), start= c(Ka1=0.0001,Ka2=0.01))
tryfit
## Nonlinear regression model
##   model: tf ~ (H/Ka1 + 2 * H^2/(Ka1 * Ka2))/(1 + H/Ka1 + H^2/(Ka1 * Ka2))
##    data: parent.frame()
##       Ka1       Ka2 
## 7.105e-07 1.517e-02 
##  residual sum-of-squares: 0.06941
## 
## Number of iterations to convergence: 6 
## Achieved convergence tolerance: 8.707e-06
summary(tryfit)
## 
## Formula: tf ~ (H/Ka1 + 2 * H^2/(Ka1 * Ka2))/(1 + H/Ka1 + H^2/(Ka1 * Ka2))
## 
## Parameters:
##      Estimate Std. Error t value Pr(>|t|)    
## Ka1 7.105e-07  1.841e-08   38.59   <2e-16 ***
## Ka2 1.517e-02  3.579e-04   42.38   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.02945 on 80 degrees of freedom
## 
## Number of iterations to convergence: 6 
## Achieved convergence tolerance: 8.707e-06
lines(pHVect,predict(tryfit),col='blue')

R gave us the values Ka2= 7.105e-07 and Ka1= 1.517e-02.

Discussion

R gave us the Ka1=1.517e-02 and from the lab report our Ka1 was 6.457e-03, this is an error of (((1.517e-02-6.457e-03)/1.517e-02)100= 57.44%). That’s a large percentage error. The Ka2 that R calculated was 7.105e-07 and from the lab report our Ka2 was 1.380e-05, this error percentage is (((7.105e-07-1.380e-05)/7.105e-07)100= 1842%). This is an even greater error which is due to the fact that when you find the equivalence point and then try to go to the half equivalence point the data is limited and can’t give you the exact number so you have to find the nearest number which is sometimes really far. I believe that the R version is better simply because it is using a best fit line and I wonder had we done this manually would our results be this high in error percentage.

References

Reference #1:Potentiometric Titrators | Thermo Fisher Scientific - IE. (n.d.-a). https://www.thermofisher.com/us/en/home/life-science/lab-equipment/ph-electrochemistry/potentiometric-titrators.html

Reference #2: Potentiometric Titrators | Thermo Fisher Scientific - IE. (n.d.-b). https://www.thermofisher.com/us/en/home/life-science/lab-equipment/ph-electrochemistry/potentiometric-titrators.html#:~:text=An%20automatic%20titrator%20allows%20you,any%20involvement%20from%20the%20operator