Question
I need help to with these lap questions plz
LABORATORY EXPERIMENT 6 TITRATION Precision of measurement is one of the keystones of physical science. The instrument most frequently employed to measure amounts of substances involved in a reaction is a balance. The degree of precision depends on the type of balance used. The most precise (and naturally the most costly ones) are the analytical balances, which can render measurement with a precision of one part in one thousand. Analytical balances, though costly and sensitive, are very well suited to the measurements of solids. As most chemical reactions, however, take place in solution, an alternate, equally precise and less costly procedure has been devised for the measurement of amounts of substances in solution. This procedure is called titration, and the basic component is a calibrated St delivery tube called a buret. The procedure itself is an exceedingly simple one. The apparatus consists of an erlenmeyer flask, buret and stand, as shown in Figure 1. A precisely measured volume of one reactant is placed in the erlenmeyer flask, and the second reactant is delivered from the buret in small increments until the reaction has gone to completion, the equivalence point. ank One of the most common and familiar reactions in chemistry is the reaction of an acid with a base. This reaction is called neutralization, and the essential feature of this process in aqueous solution is the combination of hydronium ions with hydroxide ions to form water H30"(aq) + OH (aq) 2H20(1) In this experiment you will use this reaction to determine accurately the concentration of a sodium hydroxide solution. A solution of sodium hydroxide will be standardized by titrating an oxalic acid solution of known concentration against it Page 1
Explanation / Answer
Q1.
titration technique, which allows us to relate analytically the concentration of an unkown
Q2
a
because the buret is marked so we just need to record initial volume and final volume, with this, the TOTAL volume can be claculated with a simple substraction
Vused = Vfinal - Vinitial
b
Yes, you can apply reverse titration, that is, titrate the "unkown" extra acid with another set of experiment, using the same solution
Q3
m = 0.063 g of acid
M = mol/V
MW of hdyrated acid = 90.03 + 2*18 = 126.03 g/mol
mol = mass/MW = (0.063) / (126.03) = 0.000499
M = mol/V = (0.000499)/(0.25) = 0.001996 M
Q4
M = mol/V
mol of base =mass/MW = 0.212/105.9884 = 0.0020
V = 24.5*10^-3
M = (0.0020) / (24.5*10^-3) = 0.0816 M
Q5
sources --> incorrect manipulation of buret; not adding enough indicator, incorrect measurment of acid, forgetting the hydrate presence in the acid formula
