An unknown concentration of an HC1 solution is to be determined by titration of
ID: 530844 • Letter: A
Question
An unknown concentration of an HC1 solution is to be determined by titration of known amounts of THAM. A 25.03-mL aliquot of the original unknown HC1 solution is diluted with deionized water to a volume of 249.7 mL using the calibrated volumes of volumetric pipette and volumetric flask, respectively. This solution was then mixed well and used to fill the buret. For the first titration, 0.3079 g of THAM was weighed out and dissolved in 50 mL of deionized water, and Bromocresol green indicator was added. A distinct color - change was noticed after 22.71 mL of the HC1 titrant had been added to the solution. The next two titrations were similarly carried out with 23.00 mL of HC1 being required to titrate 0.3124 g of THAM for the second trial, and 22.05 mL of HC1 reacting with 0.3005 g of THAM for the third. Use these results to calculate three determinations of the concentration of the original unknown solution. Calculate and report the average and standard deviation of the individual results.Explanation / Answer
Let x mol/L be the concentration of the original unknown HCl solution.
Moles of HCl taken = (25.03 mL)*( x mol/L) = 25.03x mmole.
The HCl solution was diluted to 249.7 mL; therefore, concentration of HCl in the dilute solution = (25.03x mmole)/(249.7 mL) = (25.03x/249.7) M.
22.71 mL of (25.03x/249.7) M HCl was required for the first trial. Therefore, the moles of HCl neutralized = (22.71 mL)*(25.03x/249.7) mol/L = (22.71*25.03x/249.7) mmole.
THAM is [(HOCH2)3NH2] that reacts with HCl in a 1:1 ratio as below:
(HOCH2)3NH2 + HCl -----> (HOCH2)3NH3+Cl-
Molar mass of THAM = 121.14 g/mol.
Mass of THAM taken in the first trial = (0.3079 g)/(121.14 g/mol) = 2.5417*10-3 mole = (2.5417*10-3 mole)*(1000 mmole/1 mole) = 2.5417 mmole.
Since THAM reacts with HCl in a 1:1 molar ratio, we must have
2.5417 = 22.71*25.03x/249.7
===> x = 2.5417*249.7/(22.71*25.03) = 1.1165
The concentration of HCl as per the first trial is 1.1165 mol/L.
Second Trial:
Moles of THAM taken = (0.3124 g)/(121.14 g/mol) = 2.5788*10-3 mole = (2.5788*10-3 mole)*(1000 mmole/1 mole) = 2.5788 mmole.
As per the problem,
2.5788 = 23.00*23.05x/249.7
===> x = 2.5788*249.7/(23.00*23.05) = 1.2146
The concentration of HCl as per the second trial is 1.2146 mol/L.
Third Trial:
Moles of THAM taken = (0.3005 g)/(121.14 g/mol) = 2.4806*10-3 mole = (2.4806*10-3 mole)*(1000 mmole/1 mole) = 2.4806 mmole.
As per the problem,
2.4806 = 22.05*23.05x/249.7
===> x = 2.4806*249.7/(22.05*23.05) = 1.2187
The concentration of HCl as per the third trial = 1.2187 mol/L.
The average concentration of HCl = 1/3*(1.1165 + 1.2146 + 1.2187) mol/L = 1.1833 mol/L (ans).
The standard deviation of the average = ?[{(1.1165 – 1.1833)2 + (1.2146 – 1.1833)2 + (1.2187 – 1.1833)2}/3] mol/L = 0.047 mol/L (ans).
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.