Question

help please! thank you!

2. In the Spectrophotometric analysis of your unknown compound you will create a standard curve with a stock solution of 0.400 M Cu. Determine the concentration of Cu in tubes 1 -4. Complete the values here and in this same table in your lab book. tube Contents Concentration of Cu ions (M Approximately 10 mL of1 M HNO 4.0 mL of 0.400 M Cu (stock standard) and 6.0 mL 1 M HNO 2 7.0 mL of 0.400 M Cu (stock standard) and 3.0 mL 1 M HNO Approximately 10 mL of 0.400 M Cu stock standard) 4

Answer 1

we have to depict the concentration of Cu²⁺ in each case. Before proceeding ahead, we have to always keep one thing in mind that while adding the solutions, the total volume always gets added up.Although it seems pretty primitive, but we have to take care of it!! So now let's move ahead.

Let's assume the volume of the stock solution to be 1L = 1000mL

1) 10 ml of HNO₃ , so here we don't have the addition of any Cu²⁺ ion but only the total volume increases by 10 ml. Therefore, dealing in millimoles, we have concentration of Cu²⁺ as

1000 * 0.400 = M * (1000+10), Therefore M = 0.4 * 100/101 = 0.396

Therefore, concentration of Cu²⁺ ions = 0.396 M

2) Applying the same concept here, we have M = (4 * 0.4 + 1000 * 0.4)/(1000+4+6) = 0.398

Therefore, concentration of Cu²⁺ ions = 0.398 M

3) Similar to previous problem, we have M = (7 * 0.4 + 1000 * 0.4)/(1000 + 7 +3) = 0.399

Therefore, concentration of Cu²⁺ ions = 0.399 M

4) This one we can predict even without carrying out any calculations, its 0.4 M. That's clear because the added solution also has the same concentration as the stock solution.

*All the best*