Question

1) (a)Which will have greater absorbance as measured by a spectrometer: weak ice tea or strong ice tea?
(b)Which will have a greater transmittance?(Hint: see the Appendix)
2) Why do we usually use the wavelength of maximum absorbance to make quantative measurements of concentration?
3) What is the purpose of adding ammonium hydroxide to the aqueos solution of copper ions?
4) (a)What mass of CuSO4*5H2O is needed to obtain 0.100 and 0.0100g Cu^2+?
(b) if you separately dissolve each of these masses into 25.0 mL, then dilute 10.00 mL of each one to a final volume of 100.00 mL, what is the concentration in g Cu^2+ per 250.00 mL in each diluted solution?

Answer 1

a) (a) absorbance measurements are an direct measure of concentration of solution. Thus, strong ice tea with more dissolved tea in solution will give greater absorbance then weak ice tea with lower amount of tea dissolved in solution.

(b) Transmittance is opposite of absorbance, thus greater the absorbance lower would be the transmittance. In the above case, the weak tea will have greater transmittance then strong tea.

2) To have the minimum interference we use wavelength at maximum absorbance for concentration calculation. It also gives the best result and minimises other errors.

3) Ammonium hydroxide is a base which reacts with aqueous copper ions in solution to give a precipitate of copper hydroxide. The copper as such in aqueous medium exists as [Cu(H2O)6]2+ ion which is light blue in color, adding base helps remove hydrogen from the complex and form a neutral complex [Cu(H2O)4(OH)2] which is dark blue in color and easy to be used for absorbance studies.

s) (a) 1 mole of CuSO4.5H2O has 1 mole of Cu2+ in it

or, 249.6850 g CuSO4.5H2O has 63.546 g of Cu2+

so, to get 0.100 g of Cu2+ we would need = 0.100 x 249.6850 / 63.546 = 0.393 g of CuSO4.5H2O

and, for 0.0100 g of Cu2+ we would need = 0.0100 x 249.6850 / 63.546 = 0.0393 g of CuSO4.5H2O

(b) we have 0.1 g of Cu2+ in 25 mL

molarity = 0.1/249.6850 x 0.025 = 0.016 M in 25 mL

to get a 100 mL solution,

M1V1 = M2V2

0.016 x 10 = M2 x 100

M2 = 0.0016 M in 100 mL

So concentration in 250 mL = 0.0016 x 250/100 = 4.0 x 10^-3 M

For 0.01 g of Cu2+

In 25 mL molarity = 0.01/249.6850 x 0.025 = 0.0016 M

In 100 mL = 0.0016 x 10/100 = 0.00016 M

In 250 mL concentration would be = 0.00016 x 250 / 100 = 4.0 x 10^-4 M