Question

A solution is prepared by pipetting 5.00 mL of 0.0983 molar HCl into a 100.0 mL graduated cylinder, and adding enough water to prepare 50.0 mL of solution.
1.The concentration of this solution is ____________M
2.The pH of this solution is ____________

Because HCl is a strong acid therefore it will dissolve 100% therefore H=0.0983M and Cl =0.0983
pH=-log(0.0983)
I have this but it not correct.

Answer 1

(1)Ca. Moles H = Molarity x Volume(L) = 0.005L(0.0983M)HCl = 0.00049 mole HCl in 50ml of solution. Now, covert to [HCl] = (moles HCl)/(Vol. Soln in Liters) = 0.00049mole/0.050L =0.0098M in HCl => [H]=0.0098M.
(2) pH = -log[H] = -log(0.0098) = 2.01