A solution is prepared by mixing 5.00 mL of 0.100 M HCl and 2.00 mL of 0.200 M NaCl. What is the molarity of chloride ion in this solution?

0.129 M

Explanation:

0.100 M HCl = 0.100 mol/L solution HCl

5.00 mL = 0.00500 L solution HCl

0.100 mol/L HCl * 0.00500 L = 0.000500 mol HCl

                             HCl > H+ + Cl-

                           1 mol                   1 mol

                    0.000500 mol           0.000500 mol

0.200 M NaCl = 0.200 mol/L solution NaCl

2.00 mL = 0.00200 L solution NaCl

0.200 mol/L NaCl*0.00200 L = 0.000400 mol NaCl

                              NaCl > Na+ + Cl-

                            1 mol                        1 mol

                     0.000400 mol               0.000400 mol

Chloride ion altogether (0.000500 mol + 0.000400 mol) =0.000900 mol

Solution altogether (0.00500 L+0.00200 L) = 0.00700L

Molarity (Cl-)= solute/solution = 0.000900 mol/0.00700L = 0.129 mol/L=

= 0.129 M

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