Prepare theoretical titration curves for the following two titrations. I each ca
ID: 917143 • Letter: P
Question
Prepare theoretical titration curves for the following two titrations. I each case, calculate the pH at the 6 points of the titration( fraction titrated=0, 0.25, 0.50, 0.75, 1.0, 1.25)
a) 20.0ml of 0.20M sodium cyanide, NaCN, titrated with 0.20M HCl. Ka for hydrocyanic acid , HCN is 6.2x10-10.
b) 15.0ml of a 0.10M solution of cyanic acid, HOCN, titrated with 0.075M NaOH. Ka for cyanic acid is 3.5x10-4. Hint: since these concentrations are NOT the same, first calculate the volume of NaOH required to reach the equivalence point.
Plese help
Explanation / Answer
Ka = 6.2 x 10^-10
Kb = 1.61 x 10^-5
pKa = 9.21
pKb = 4.79
a ) o ml HCL added
here only salt NaCN reamins
pH = 7 + 1/2 [pKa + logC]
pH = 7 + 1/2 [9.21 + log0.2]
pH = 11.26
b) 0.25 ml HCl added
millimoles of HCl = 0.2 x 0.25 = 0.05
millimoles of NaCN = 20 x 0.2 = 4
NaCN + HCl-----------------------> NaCl + HCN
4 0.05 0 0
3.95 0 0.05 0.05
pH = pKa + log [NaCN] / [HCN]
pH = 9.21 + log (0.05 / 3.95)
pH = 7.31
c) 0.50 ml HCl added
millimoles of HCl = 0.2 x 0.50 = 0.1
millimoles of NaCN = 20 x 0.2 = 4
NaCN + HCl-----------------------> NaCl + HCN
4 0.1 0 0
3.9 0 0.1 0.1
pH = pKa + log [NaCN] / [HCN]
pH = 9.21 + log (0.1 / 3.9)
pH = 7.62
d) 0.75 ml HCl added
millimoles of HCl = 0.2 x 0.75 = 0.15
millimoles of NaCN = 20 x 0.2 = 4
NaCN + HCl-----------------------> NaCl + HCN
4 0.15 0 0
3.85 0 0.15 0.15
pH = pKa + log [NaCN] / [HCN]
pH = 9.21 + log (0.15 / 3.85)
pH = 7.80
e) 1.0 ml HCl added
millimoles of HCl = 0.2 x 1 = 0.2
millimoles of NaCN = 20 x 0.2 = 4
NaCN + HCl-----------------------> NaCl + HCN
4 0.2 0 0
3.8 0 0.2 0.2
pH = pKa + log [NaCN] / [HCN]
pH = 9.21 + log (0.2 / 3.8)
pH = 7.93
f) 1.25 ml HCl added
millimoles of HCl = 0.2 x 1.25 = 0.25
millimoles of NaCN = 20 x 0.2 = 4
NaCN + HCl-----------------------> NaCl + HCN
4 0.25 0 0
3.75 0 0.25 0.25
pH = pKa + log [NaCN] / [HCN]
pH = 9.21 + log (0.25 / 3.75)
pH = 8.03