Experiment 5 Prelaboratory Assignment: A Volumetric Analysis
a. The analyze is the substance n a titration whose concentration is unknown.
b. The analyze is generally added to the tyrant, not the indicator.
a. a primary standard is a substance that has a hkown high degree of purity and relatively large molar mass, is non hygroscopic and reacts in a predicable way.
b. Secondary standards are compounds used in alaysis after evaluation against primary standard. NaOH.
Stoichiometric point is when amount of moles in both your solutions are the same.
The endpoint is a visual indicator, brought about when the indicator changes color when the all the cid is neutralized, etc. when the titration process is complete.
a. If after the final rinse with ionized water, no water droplets adhere to the clean part of the glassware.
b. It should be dispensed through the buret tip or the top opening of the buret.
c. NaOh will eliminate and miniime contaminants and error %. Water has an affinity to glass. If you have water in your buret at certain level that you want it filled with to, but with NaOH. By rinsing the buret with NaOh you are ridding the buret of water left over from rinsing, creating working concentration closer to that of whci h you have calculated.
d. carefully opened the stopcock just batrelyt and let a drop grow on the tip, but don’t let it get too large to the point where it drops off. Touch the tiop of the buret to the side wall of your flask and then wash the drop into the flask with water. Also see Technique 16C
1. A 4 g mass of NaOH is dissolved in 5mL of water
a. what is the approximate molar concentration of the NaOH?
4 g X 1 mol NaOH/40.00 g NaOH = .1 mol NaOH
M=m/V .1 mol NaOH/.005L = 20 M
b. a 4 mL aliquot of this solution is diluted to 500 mL of solution. What is the approximate molar concentration of NaOH in the diluted solution?
m=M X V
20M X .004 = .08 m NaOH
M= .08m/.5L = .16 M
c. Calculate the mass of KHC8H4O4 that reacts with 15 mL of the NaOH solution
KHC8H4O4 + NaOH ----- H2O + NaKC8H4O4
15 mL/1000 X .16 M NaOH/ 1L X 1 mol KHC8H4O4/1 mol NaOH X 204.44 g KHC8H4O4/ 1 mol KHC8H4O4 = 0.49 g KHC8H4O4
a. a. KHP + NaOH ==> H2O + NaKP
moles KHP = 0.411 g / 204.44 g/mole = 0.00201 moles KHP = moles NaOH since they react in a 1:1 mole ratio.
moles NaOH = M NaOH x L NaOH
0.00201 = M NaOH x 0.01517
M NaOH = 0.132 M
b. b. HNO3 + NaOH ==> H2O + NaNO3
moles NaOH = M NaOH x L NaOH = (0.132)(0.01677) = 0.00221364 moles NaOH = moles HNO3 since they react in a 1:1 mole ratio.
moles HNO3 = M HNO3 x L HNO3
0.00221364 = M HNO3 x 0.02500
M HNO3 = 0.05299593009