Question

what volume of 6M HCl must be added to a 2mL of solution initially containing no Cl- ions so that the final CL- ion concentration will be .03 M afterwards?

Answer 1

We now that the HCL is a strong acid so it will dissociate completly to form H+ and Cl- .

HCL H+ + Cl-

1 mol of HCL will produce 1 mol of Cl-

So the number of mol will be the same.

So you want 0.03 M of HCl
x = amount of 6M of HCl
(We'll use L and scale it back to mL later, makes calculation easier)
x * 6 = 0.03 (0.002 + x) = 0.00006 + 0.03x = 6x
0.00006 = 6x-0.003x

0.00006 = 5.997x

x= 0.000010 L = 0.01 mL of 6M HCl