Binding
Kengio Chan
2024-03-27
Abstract (about comaring titration analysis versus binding analysis)
As more and more NaOH being added to the solution, the pH value increase.
Experimental
Introduction (about prior work and purpose of paper)
The titration curve and the binding curve of monoprotic acid of trial 2 are being compared. The binding curve of diprotic acid of trial 3 is being discussed.
The purpose of the paper is to compare the titration curve and the binding curve formed by using the monoprotic acid data from the lab. The diprotic acid data is also used to form a binding curve to compare its KD values with the KD value of the monoprotic acid data.
Titration Analysis for monoprotic acid
\(F = \frac {H}{H+KD}\)
Mydata <- read.csv("mydata.csv")
Mydata## Volume..ml...measured. pH..measured.
## 1 0.0 3.130
## 2 1.0 3.371
## 3 2.0 3.579
## 4 3.0 3.757
## 5 4.0 3.909
## 6 5.0 4.032
## 7 6.0 4.150
## 8 7.0 4.257
## 9 8.0 4.355
## 10 9.0 4.445
## 11 10.0 4.505
## 12 11.0 4.628
## 13 12.0 4.710
## 14 13.0 4.813
## 15 14.0 4.914
## 16 15.0 5.033
## 17 16.0 5.165
## 18 16.5 5.293
## 19 17.0 5.354
## 20 17.5 5.436
## 21 18.0 5.556
## 22 18.5 5.715
## 23 19.0 5.928
## 24 19.5 6.255
## 25 20.0 7.019
## 26 20.5 9.452
## 27 21.0 10.181
## 28 21.5 10.551
## 29 22.0 10.779
## 30 22.5 10.929
## 31 23.0 11.029
## 32 23.5 11.137
## 33 24.0 11.180
## 34 24.5 11.230
## 35 25.0 11.282
## 36 25.5 11.330
## 37 26.0 11.368
## 38 26.5 11.402
## 39 27.0 11.432
## 40 27.5 11.464
## 41 28.0 11.488
## 42 28.5 11.511
## 43 29.0 11.534
## 44 29.5 11.559
## 45 30.0 11.583
## 46 31.0 11.608
## 47 32.0 11.635
## 48 33.0 11.660
## 49 34.0 11.683
## 50 35.0 11.706
## 51 36.0 11.728
## 52 37.0 11.749
## 53 38.0 11.763
## 54 39.0 11.783
## 55 40.0 11.797
## 56 41.0 11.814
## 57 42.0 11.828
## 58 43.0 11.846
## 59 44.0 11.856
## 60 45.0 11.869
## 61 46.0 11.882
## 62 47.0 11.887
## 63 48.0 11.901
## 64 49.0 11.908
## 65 50.0 11.923
## 66 51.0 11.930
## 67 52.0 11.937
## 68 53.0 11.942volume <- Mydata$Volume..ml...measured.
volume## [1] 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 14.0
## [16] 15.0 16.0 16.5 17.0 17.5 18.0 18.5 19.0 19.5 20.0 20.5 21.0 21.5 22.0 22.5
## [31] 23.0 23.5 24.0 24.5 25.0 25.5 26.0 26.5 27.0 27.5 28.0 28.5 29.0 29.5 30.0
## [46] 31.0 32.0 33.0 34.0 35.0 36.0 37.0 38.0 39.0 40.0 41.0 42.0 43.0 44.0 45.0
## [61] 46.0 47.0 48.0 49.0 50.0 51.0 52.0 53.0pH <- Mydata$pH..measured.
pH## [1] 3.130 3.371 3.579 3.757 3.909 4.032 4.150 4.257 4.355 4.445
## [11] 4.505 4.628 4.710 4.813 4.914 5.033 5.165 5.293 5.354 5.436
## [21] 5.556 5.715 5.928 6.255 7.019 9.452 10.181 10.551 10.779 10.929
## [31] 11.029 11.137 11.180 11.230 11.282 11.330 11.368 11.402 11.432 11.464
## [41] 11.488 11.511 11.534 11.559 11.583 11.608 11.635 11.660 11.683 11.706
## [51] 11.728 11.749 11.763 11.783 11.797 11.814 11.828 11.846 11.856 11.869
## [61] 11.882 11.887 11.901 11.908 11.923 11.930 11.937 11.942plot(volume,pH, main="Titration curve of trial 2",
xlab="Volume of NaOH in ml", ylab= "pH",
xlim = c(0.0, 55.0),ylim = c(0.0, 14.0))
From the titration curve, it shows an equivalence point on the graph.As the volume of NaOH increase, the pH value also increase. pH value gets larger when there is more NaOH being added to the solution.
Binding Curve Transformation
H <- 10^-(pH) # H+
tbase <- 0.10
tbase## [1] 0.1# endpoint volume
VE <- 20.0
# volume added:
VA <- volume
# initial volume:
VI <- 26.20
# the concentration of the base NaOH:
CB <- 0.0962
fb <- (1-(((VA*CB)+((H)*(VI+VA)))/(VE*CB)))
fb## [1] 9.899052e-01 9.439832e-01 8.961359e-01 8.473443e-01 7.980645e-01
## [6] 7.484936e-01 6.988152e-01 6.490452e-01 5.992151e-01 5.493433e-01
## [11] 4.994118e-01 4.495447e-01 3.996129e-01 3.496866e-01 2.997453e-01
## [16] 2.498015e-01 1.998500e-01 1.748870e-01 1.499006e-01 1.249168e-01
## [21] 9.993614e-02 7.495522e-02 4.997227e-02 2.498680e-02 -2.298460e-06
## [26] -2.500001e-02 -5.000000e-02 -7.500000e-02 -1.000000e-01 -1.250000e-01
## [31] -1.500000e-01 -1.750000e-01 -2.000000e-01 -2.250000e-01 -2.500000e-01
## [36] -2.750000e-01 -3.000000e-01 -3.250000e-01 -3.500000e-01 -3.750000e-01
## [41] -4.000000e-01 -4.250000e-01 -4.500000e-01 -4.750000e-01 -5.000000e-01
## [46] -5.500000e-01 -6.000000e-01 -6.500000e-01 -7.000000e-01 -7.500000e-01
## [51] -8.000000e-01 -8.500000e-01 -9.000000e-01 -9.500000e-01 -1.000000e+00
## [56] -1.050000e+00 -1.100000e+00 -1.150000e+00 -1.200000e+00 -1.250000e+00
## [61] -1.300000e+00 -1.350000e+00 -1.400000e+00 -1.450000e+00 -1.500000e+00
## [66] -1.550000e+00 -1.600000e+00 -1.650000e+00plot(pH,fb, main="Binding curve of trial 2",
xlab="Volume of NaOH in ml", ylab= "fb",
xlim = c(2.5,10.0), ylim =c(0.0,1.0))
library(nls2)## Loading required package: protofit <- nls2(fb ~ H/(KD+H), start = c(KD=0.001))
summary (fit)##
## Formula: fb ~ H/(KD + H)
##
## Parameters:
## Estimate Std. Error t value Pr(>|t|)
## KD 2.910e-05 2.444e-05 1.191 0.238
##
## Residual standard error: 0.6928 on 67 degrees of freedom
##
## Number of iterations to convergence: 6
## Achieved convergence tolerance: 7.899e-07lines(pH, predict(fit), col="red")
The data was transformed to binding curve by using the data.The data is being plugged into the monoprotic acid’s fraction bound equation to get the fraction bound values. The binding curves is formed. The volumes of NaOH added to the solution as the x-axis and the fraction bound values as the y-axis.
By using nls2, the KD1 value is being calculated. From the equation, \(F = \frac {H}{H+KD}\), the value of KD1 is 2.910e-05.
From the regular lab report of monoprotic acid, the KD value of monoprotic acid is 3.03e-08.
The KD value from the regular lab report tends to be smaller than the KD1 value calculated from the equation.
When comparing to the traditional titration analysis to a binding curve, the pros are the binding curve could help to find the value of equilibrium constant and the titration analysis is accurate. The titration curve could also help to find the equivalence point.
The cons of the titration analysis are it takes a longer period of time to record the data and large amount of data needed to be recorded. A binding curve also need large amount of data to form the curve, which takes a longer period of time.
Diprotic data recorded for trial 3
\(F =\frac{\frac{H}{KD1}+\frac{2H^2}{KD1*KD2}}{1+{\frac{H}{KD1}+\frac{H^2}{KD1*KD2}}}\)
# Reading the diprotic data:
pH <-read.csv("diproticpHdata.csv")
pH## Potential..pH.
## 1 1.84
## 2 1.84
## 3 1.85
## 4 1.85
## 5 1.85
## 6 1.85
## 7 1.86
## 8 1.88
## 9 1.89
## 10 1.91
## 11 1.93
## 12 1.94
## 13 1.96
## 14 1.99
## 15 2.02
## 16 2.04
## 17 2.06
## 18 2.08
## 19 2.11
## 20 2.14
## 21 2.18
## 22 2.19
## 23 2.24
## 24 2.26
## 25 2.30
## 26 2.33
## 27 2.37
## 28 2.42
## 29 2.47
## 30 2.54
## 31 2.60
## 32 2.64
## 33 2.70
## 34 2.82
## 35 2.91
## 36 3.04
## 37 3.23
## 38 3.48
## 39 3.65
## 40 3.84
## 41 4.00
## 42 4.16
## 43 4.34
## 44 4.55
## 45 4.67
## 46 4.86
## 47 5.05
## 48 5.18
## 49 5.28
## 50 5.39
## 51 5.48
## 52 5.57
## 53 5.63
## 54 5.69
## 55 5.75
## 56 5.81
## 57 5.82
## 58 5.92
## 59 5.96
## 60 5.97
## 61 6.07
## 62 6.10
## 63 6.15
## 64 6.19
## 65 6.23
## 66 6.28
## 67 6.33
## 68 6.38
## 69 6.43
## 70 6.44
## 71 6.53
## 72 6.58
## 73 6.63
## 74 6.71
## 75 6.78
## 76 6.78
## 77 6.90
## 78 7.00
## 79 7.01
## 80 7.18
## 81 7.20
## 82 7.49
## 83 7.60
## 84 7.61
## 85 8.22
## 86 8.41
## 87 8.44
## 88 8.46
## 89 8.54
## 90 8.75
## 91 8.81
## 92 8.93
## 93 9.12
## 94 9.27
## 95 9.49
## 96 9.71
## 97 9.83Vol <- read.csv("diproticconsumptiondata.csv")
Vol## Consumption..mL.
## 1 0.000
## 2 0.005
## 3 0.010
## 4 0.023
## 5 0.054
## 6 0.132
## 7 0.328
## 8 0.528
## 9 0.728
## 10 0.928
## 11 1.128
## 12 1.328
## 13 1.528
## 14 1.728
## 15 1.928
## 16 2.128
## 17 2.328
## 18 2.528
## 19 2.728
## 20 2.928
## 21 3.129
## 22 3.329
## 23 3.529
## 24 3.729
## 25 3.929
## 26 4.129
## 27 4.329
## 28 4.529
## 29 4.729
## 30 4.929
## 31 5.129
## 32 5.329
## 33 5.529
## 34 5.729
## 35 5.929
## 36 6.129
## 37 6.329
## 38 6.529
## 39 6.639
## 40 6.733
## 41 6.807
## 42 6.885
## 43 6.977
## 44 7.078
## 45 7.163
## 46 7.336
## 47 7.536
## 48 7.736
## 49 7.936
## 50 8.136
## 51 8.337
## 52 8.537
## 53 8.737
## 54 8.937
## 55 9.137
## 56 9.337
## 57 9.537
## 58 9.737
## 59 9.937
## 60 10.137
## 61 10.337
## 62 10.537
## 63 10.737
## 64 10.937
## 65 11.137
## 66 11.337
## 67 11.537
## 68 11.737
## 69 11.937
## 70 12.137
## 71 12.337
## 72 12.537
## 73 12.737
## 74 12.937
## 75 13.137
## 76 13.337
## 77 13.537
## 78 13.737
## 79 13.937
## 80 14.137
## 81 14.337
## 82 14.537
## 83 14.615
## 84 14.775
## 85 14.975
## 86 14.993
## 87 14.998
## 88 15.012
## 89 15.044
## 90 15.082
## 91 15.097
## 92 15.137
## 93 15.213
## 94 15.303
## 95 15.467
## 96 15.639
## 97 15.799length(Vol)## [1] 1VolVect <-Vol$Consumption..mL.
VolVect## [1] 0.000 0.005 0.010 0.023 0.054 0.132 0.328 0.528 0.728 0.928
## [11] 1.128 1.328 1.528 1.728 1.928 2.128 2.328 2.528 2.728 2.928
## [21] 3.129 3.329 3.529 3.729 3.929 4.129 4.329 4.529 4.729 4.929
## [31] 5.129 5.329 5.529 5.729 5.929 6.129 6.329 6.529 6.639 6.733
## [41] 6.807 6.885 6.977 7.078 7.163 7.336 7.536 7.736 7.936 8.136
## [51] 8.337 8.537 8.737 8.937 9.137 9.337 9.537 9.737 9.937 10.137
## [61] 10.337 10.537 10.737 10.937 11.137 11.337 11.537 11.737 11.937 12.137
## [71] 12.337 12.537 12.737 12.937 13.137 13.337 13.537 13.737 13.937 14.137
## [81] 14.337 14.537 14.615 14.775 14.975 14.993 14.998 15.012 15.044 15.082
## [91] 15.097 15.137 15.213 15.303 15.467 15.639 15.799length(VolVect)## [1] 97pHVect <- pH$Potential..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 7.60 7.61 8.22 8.41 8.44 8.46 8.54 8.75
## [91] 8.81 8.93 9.12 9.27 9.49 9.71 9.83length(pHVect)## [1] 97H <- 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 2.511886e-08 2.454709e-08 6.025596e-09
## [86] 3.890451e-09 3.630781e-09 3.467369e-09 2.884032e-09 1.778279e-09
## [91] 1.548817e-09 1.174898e-09 7.585776e-10 5.370318e-10 3.235937e-10
## [96] 1.949845e-10 1.479108e-10#Volume added at the 1st endpoint:
VE=13.76/2
# Initial volume of unknown acid:
VI=25.00
# The concentratiion of the base NaOH:
CB=0.10
CB## [1] 0.1VA <- VolVect
F <- (2-(((VA*CB)+((H)*(VI+VA)))/(VE*CB)))
fb <- F
fb## [1] 1.474767525 1.473935734 1.485064477 1.482908038 1.477765761
## [6] 1.464827128 1.444151560 1.434121250 1.412440633 1.401476618
## [11] 1.389858673 1.367607725 1.355125384 1.351299781 1.345987786
## [16] 1.331089680 1.315673067 1.299755914 1.290642732 1.280348147
## [21] 1.275077820 1.250279961 1.248448750 1.228520934 1.218185385
## [26] 1.201821847 1.188936763 1.178537606 1.166228228 1.158116019
## [31] 1.144504904 1.124448074 1.107829414 1.099694442 1.082920083
## [36] 1.067892463 1.053273426 1.035842684 1.024733881 1.014699398
## [41] 1.005987355 0.996067001 0.983776699 0.969906859 0.957866813
## [46] 0.933072150 0.904229684 0.875267028 0.846260392 0.817245655
## [51] 0.788066295 0.759025776 0.729972257 0.700916729 0.671859440
## [56] 0.642800608 0.613732160 0.584677670 0.555612925 0.526544113
## [61] 0.497485354 0.468418273 0.439352762 0.410286042 0.381219071
## [66] 0.352152515 0.323085626 0.294018438 0.264950983 0.235881564
## [71] 0.206815379 0.177747277 0.148679002 0.119611341 0.090543126
## [76] 0.061473311 0.032405739 0.003337393 -0.025732275 -0.054800270
## [81] -0.083869887 -0.112937906 -0.124274702 -0.147530489 -0.176599187
## [86] -0.179215342 -0.179942072 -0.181976946 -0.186628075 -0.192151266
## [91] -0.194331486 -0.200145417 -0.211191905 -0.224273287 -0.248110484
## [96] -0.273110477 -0.296366288length(pHVect)## [1] 97plot(pHVect,fb)
library(nls2)
fit<- nls2(fb ~ (H/KD1+2*H^2/(KD1*KD2))/(1+H/KD1+H^2/(KD1*KD2)),
start = c(KD1=0.0001,KD2=0.01))
fit## Nonlinear regression model
## model: fb ~ (H/KD1 + 2 * H^2/(KD1 * KD2))/(1 + H/KD1 + H^2/(KD1 * KD2))
## data: parent.frame()
## KD1 KD2
## 1.003e-06 1.748e-02
## residual sum-of-squares: 0.7804
##
## Number of iterations to convergence: 8
## Achieved convergence tolerance: 4.015e-06summary (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 1.003e-06 8.093e-08 12.39 <2e-16 ***
## KD2 1.748e-02 1.317e-03 13.28 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.09064 on 95 degrees of freedom
##
## Number of iterations to convergence: 8
## Achieved convergence tolerance: 4.015e-06length((predict(fit)))## [1] 97plot(pHVect,fb, main="Binding curve of trial 3",
xlab="pH", ylab= "fb",
xlim = c(1.5,7.0),ylim = c(0.0, 1.6))
lines(pHVect,predict(fit),col="red")
The fraction bound equation for diprotic acid is being used to find the fraction bound values.
The data recorded from the lab could be plugged into the equation, \(F =\frac{\frac{H}{KD1}+\frac{2H^2}{KD1*KD2}}{1+{\frac{H}{KD1}+\frac{H^2}{KD1*KD2}}}\), to find the values of Ka1 and Ka2.
By using nls2 and the equation above, the value of Ka1 is 1.003e-06 and the value of Ka2 is 1.748e-02.
From the regular lab report of monoprotic acid, the Ka value of monoprotic acid is 3.03e-08. From the regular lab report of diprotic acid, the Ka1 value is 5.25e-06 and Ka2 is 4.47e-09.
When comparing Ka1 and Ka2 values from the regular lab report with the Ka values of the diprotic acid, the value of Ka1 tends to be larger than Ka2. Ka2 from the regular lab report is much more smaller than the Ka2 calculated from nls2 and the equation.
The future thoughts I have for the future studies are I might be adding other compounds to the solution instead of NaOH and observe how adding different compound could affect the pH value, like whether the pH value would increase or decrease as the volume added to the solution increase.
From my result, I could know that I need to repeat the procedure few more times to make sure that I get accurate results. Make sure no errors being made when calculating Ka values.