To identify the unknown diprotic acid sample, an autotitrator was used to find the equivalence points, molar mass and acid dissociation of the unknown. The acid was titrated with 0.1000M NaOH solution at 24C. the pKa No. 1 was determined by the pH at the volume of 1/2 of the equivalence point No. 1, and pKa No. 2 was determined at the volume by which (Vol. NaOH at 1st EQ + Vol. NaOH at 2nd EQ)/2
The Ka 1 and Ka 2 was determined by taking the 10^-pKa No. 1 and 10^-pKa No.2.
data <- read.csv("autotitration.csv")
VolVect <- data$mL
VolVect
## [1] 0.000 0.005 0.010 0.023 0.054 0.132 0.328 0.528 0.728 0.928
## [11] 1.128 1.328 1.528 1.728 1.928 2.128 2.328 2.528 2.728 2.928
## [21] 3.129 3.329 3.529 3.729 3.929 4.129 4.329 4.529 4.729 4.929
## [31] 5.129 5.329 5.529 5.729 5.929 6.129 6.329 6.529 6.729 6.929
## [41] 7.129 7.329 7.529 7.729 7.929 8.129 8.329 8.529 8.729 8.929
## [51] 9.129 9.329 9.529 9.730 9.930 10.130 10.330 10.530 10.730 10.930
## [61] 11.130 11.276 11.476 11.676 11.876 12.076 12.276 12.309 12.314 12.327
## [71] 12.358 12.437 12.634 12.714 12.914 12.931 12.972 12.977 12.990 13.021
## [81] 13.026 13.039 13.044 13.056 13.088 13.093 13.100 13.119 13.124 13.137
## [91] 13.142 13.154 13.159 13.172 13.203
pHVect <- data$pH
pHVect
## [1] 1.48 1.47 1.47 1.47 1.47 1.46 1.46 1.47 1.47 1.48 1.49 1.50
## [13] 1.51 1.52 1.54 1.55 1.56 1.58 1.60 1.62 1.65 1.67 1.73 1.76
## [25] 1.80 1.83 1.90 1.94 1.99 2.04 2.07 2.14 2.21 2.30 2.36 2.45
## [37] 2.60 2.66 2.73 2.86 2.97 3.02 3.14 3.23 3.28 3.35 3.40 3.48
## [49] 3.56 3.60 3.68 3.75 3.80 3.88 3.91 3.97 4.05 4.10 4.16 4.22
## [61] 4.44 4.34 4.42 4.52 4.61 4.72 5.18 4.87 4.86 4.86 4.87 4.89
## [73] 5.18 5.18 5.83 5.83 6.11 6.12 6.12 6.63 6.64 6.92 6.93 6.93
## [85] 8.77 8.96 8.97 9.56 9.57 9.78 9.80 9.96 9.97 9.98 10.36
plot(VolVect, pHVect, xlab= 'Volume (mL)', ylab= 'pH', main = "Titration curve" )
Hcon <- 10 ^(-pHVect)
Hcon
## [1] 3.311311e-02 3.388442e-02 3.388442e-02 3.388442e-02 3.388442e-02
## [6] 3.467369e-02 3.467369e-02 3.388442e-02 3.388442e-02 3.311311e-02
## [11] 3.235937e-02 3.162278e-02 3.090295e-02 3.019952e-02 2.884032e-02
## [16] 2.818383e-02 2.754229e-02 2.630268e-02 2.511886e-02 2.398833e-02
## [21] 2.238721e-02 2.137962e-02 1.862087e-02 1.737801e-02 1.584893e-02
## [26] 1.479108e-02 1.258925e-02 1.148154e-02 1.023293e-02 9.120108e-03
## [31] 8.511380e-03 7.244360e-03 6.165950e-03 5.011872e-03 4.365158e-03
## [36] 3.548134e-03 2.511886e-03 2.187762e-03 1.862087e-03 1.380384e-03
## [41] 1.071519e-03 9.549926e-04 7.244360e-04 5.888437e-04 5.248075e-04
## [46] 4.466836e-04 3.981072e-04 3.311311e-04 2.754229e-04 2.511886e-04
## [51] 2.089296e-04 1.778279e-04 1.584893e-04 1.318257e-04 1.230269e-04
## [56] 1.071519e-04 8.912509e-05 7.943282e-05 6.918310e-05 6.025596e-05
## [61] 3.630781e-05 4.570882e-05 3.801894e-05 3.019952e-05 2.454709e-05
## [66] 1.905461e-05 6.606934e-06 1.348963e-05 1.380384e-05 1.380384e-05
## [71] 1.348963e-05 1.288250e-05 6.606934e-06 6.606934e-06 1.479108e-06
## [76] 1.479108e-06 7.762471e-07 7.585776e-07 7.585776e-07 2.344229e-07
## [81] 2.290868e-07 1.202264e-07 1.174898e-07 1.174898e-07 1.698244e-09
## [86] 1.096478e-09 1.071519e-09 2.754229e-10 2.691535e-10 1.659587e-10
## [91] 1.584893e-10 1.096478e-10 1.071519e-10 1.047129e-10 4.365158e-11
\(F=2- \frac {(volume \cdot Cbase) + (Hconc \cdot (Vi + Volume))} {Vend \cdot Cbase}\)
Where: F is the fraction bound
Cbase is the concentration of NaOH
Hconc is the concentration of H+ ions
Vi is the inital volume
Vend is the volume added to the first endpoint
Vend <- 6.729
Vi <- 0.00
Vi
## [1] 0
Cbase <- 0.1
Cbase
## [1] 0.1
Fbound <- 2 - (Cbase * VolVect + Hcon * (Vi + VolVect)) / (Cbase * Vend)
Fbound
## [1] 2.00000000 1.99900517 1.99801034 1.99542378 1.98925582 1.97358162
## [7] 1.93435433 1.89494580 1.85515254 1.81642299 1.77812251 1.74023622
## [13] 1.70274972 1.66564903 1.63084540 1.59462745 1.55874804 1.52549684
## [19] 1.49275634 1.46048770 1.43089674 1.39950549 1.37789708 1.34952802
## [25] 1.32356895 1.29562731 1.27567413 1.24966581 1.22530610 1.20069399
## [31] 1.17290107 1.15068332 1.12766899 1.10593994 1.08042648 1.05684870
## [37] 1.03581850 1.00849473 0.98137913 0.95606378 0.92920365 0.90043225
## [43] 0.87300598 0.84462599 0.81548343 0.78654913 0.75729553 0.72830403
## [49] 0.69920617 0.66972379 0.64050034 0.61114734 0.58164624 0.55211374
## [55] 0.52248231 0.49296263 0.46348542 0.43388850 0.40430623 0.37470858
## [61] 0.34536468 0.32350214 0.29389760 0.26429988 0.23466857 0.20503774
## [67] 0.17553707 0.17050670 0.16975779 0.16782559 0.16322380 0.15149321
## [73] 0.12233100 0.11044137 0.08081572 0.07828931 0.07220973 0.07146702
## [79] 0.06953507 0.06493825 0.06419530 0.06226547 0.06152247 0.05973914
## [85] 0.05498585 0.05424281 0.05320254 0.05037895 0.04963590 0.04770396
## [91] 0.04696091 0.04517759 0.04443453 0.04250260 0.03789567
tF <- Fbound
plot (pHVect, tF, xlab = 'pH', ylab= 'Fraction bound', main = "Binding curve")
\(F= \frac {(Hconc/KD1 +2 \cdot Hconc^2/ (KD1 \cdot KD2))} {(1+ Hconc/KD1 + Hconc^2/ (KD1 \cdot KD2))}\)
Where:
F is the fraction bound
Hconc is the concentration of H+ ions
KD1 is the binding dissociation constant of the first proton
KD2 is the binding dissociation constant for the second proton
plot (pHVect, tF, xlab = 'pH', ylab= 'Fraction bound', main = "Binding curve")
tryfit <- nls(tF ~ (Hcon/KD1+2*Hcon^2/(KD1*KD2))/(1+Hcon/KD1+Hcon^2/(KD1*KD2)),
start =c(KD1 = 0.0001, KD2 = 0.01))
summary(tryfit)
##
## Formula: tF ~ (Hcon/KD1 + 2 * Hcon^2/(KD1 * KD2))/(1 + Hcon/KD1 + Hcon^2/(KD1 *
## KD2))
##
## Parameters:
## Estimate Std. Error t value Pr(>|t|)
## KD1 1.096e-04 1.268e-05 8.647 1.48e-13 ***
## KD2 1.639e-02 1.432e-03 11.445 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.1202 on 93 degrees of freedom
##
## Number of iterations to convergence: 6
## Achieved convergence tolerance: 4.49e-06
lines(pHVect, predict (tryfit))
Ka 1 to Ka 2 found from the nls was 1.096E^-04 and 1.639E^-02 respectively.