Question

In the laboratory a student combines 36.8 mL of a 0.252 M sodium nitrate solution with 14.4 mL of a 0.533 M copper(II) nitrate solution. What is the final concentration of nitrate anion ? 331 M Submit Answer Retry Entire Group 9 more group attempts remaining

Answer 1

Both sodium nitrate (NaNO₂) and copp- er (11) ritrate [Cu(NO3)2 are stroong electrolyte No. of moles of sodium nitrate in 36.8 mL of 0.252 M solution = 36.8 x x 0.252 molle 100 i No. of moses = 9.27 x 10-3 mol L = volm in Lx molarsity (mol/) Na Noz ² Nat & NO So, froom 9.27 x 10-3 mol of sodium nitrate we get 9. 27 x 103 mol of nitrate in . No. of moles of copper nitroate in a. 14.4 mL of 0.533 M soluti on - 14.4 x x 0.533 molst 1000 = 7.67 x 10-3 mol

Copper (11) nitrate dissociates as below - C (NO3) È Curt + 2NO3- Thus from 7-67 x 10-3 mol of C (NO3)2 we will get 2x 7.67 x 103 mol = 15.34 x 10-3 mol of NO3- ion. Thus the total no. of moles of Noz in in the solution = (9.27X10 3 + 15.34 x 103 mnder - 24.6 x 10-3 moles no. CO

Now, total volume of the solution - (36.8 + 14.4) mL = 51.2 ml = 51.2 x 103 o final concentration of nitrate anion = 24.6 x 10-3 moles 51.2x10-3 L = 0-480 mly = [0.480) Aus
I have given the answer in three significant figures.....if the answer does not match with the computer database due to significant figures.....then please comment below.......