Question

Concentration (M) of starting solution .024 1st Reading 2nd Reading Abs (15-mL .024 M Cu2+ + 7 mL conc NH3 diluted with distilled H2O to 100 mL total volume) 0.861 0.941 Abs (5-mL .024 M Cu2+ + 7 mL conc NH3 diluted with distilled H2O to 100 mL total volume) 0.541 0.518 Abs (15-mL Unknown Cu2+ solution + 7 mL conc NHz diluted with distilled H20 to 100 mL total volume) 713 713 Abs (5-mL Unknown Cu2+ solution + 7 mL conc NH3 diluted with distilled H20 to 100 mL total volume) .404 .404

BELOW I ATTACHED A OVERVIEW OF FORMULAS

SHOW ALL WORK IN DETAIL. NEED ANSWERS ASAP

Answer 1

Average of the two measurements of the 15 mL dilution sample:

0.861 +0.941 =0.901

The molarity of copper(II) of this solution can be calculated with the given formula:

C = 0.024M 15.0mL L = 0.0036 M 100ml

The M/A ratio is:

M A 15 0.0036M -= 0.0040 M 0.901

For the 5.00 mL dilution sample:

0.541 +0.518 = 0.529

The molarity of copper(II) of this solution can be calculated with the given formula:

C = 0.024M 5.00mL L = 0.0012 M 100mL

The M/A ratio is:

0.00121 -= 0.0023 M 0.529 A 5

The molarity of the 5.00 mL dilution is lower than thatof the 15.0 mL dilution.

The molarity of the unknown 15.0 mL dilution solution can be determined using the M/A ratio for the 15.0 mL sample dilution and it absorbance average value (0.713) as:

C15 = 0.0040M 0.713 = 0.0029 M

This concentration is lower than the standard 15.0 mL dilution (which was 0.0036 M).