A solution prepared by mixing 10mL of 1 M HCl and 10 mL of 1.2 M NaOH has a pH of?

Answer 1

You need to use a strong acid-strong base tritration.

NaOH + HCl ---> H2O + Na+ + Cl--

Use an ICE table to find the concentration of the base. Since you have an excess of the base (about 2mmol extra).

pOH = [OH-] = your concentration from the calculations.

pH = 14 - pOH

then you are finished

Answer 2

By mixing a solution of HCl and NaOH, you are neutralizing one or both of the solutions. In this case, since you have a greater amount of NaOH, the final solution will be alkiline. HCl + NaOH --> H2O + NaCl 10 mL of 1 M HCl = .01 mols of HCl 10 mL of 1.2 M NaOH = .012 moles of NaOH So the remaining solution (20 mL) will have .002 moles of NaOH. One can calulate the pH (or pOH) of a solution by knowing the concentration of the H+ ions (or OH- ions), pH = -log([H+]) In this case, it is easier to use pOH, pOH = -log([OH-]) pOH = -log(.002/.02) = -log(.1) = 1 since pH + pOH = 14, we know that is pOH = 1, the pH = 13

Answer 3

Since the mole ratio of HCl to NaOH in the neutralizaton reaction is 1.1, one of these is in excess. It is the NaOH. You have added .01 mole of HCl and .012 moles of NaOH. The result is an excess of 0.002 moles of NaOH in 20 mL of solution, or a 0.1 M solution. Since this means 0.1 M of [OH-] because the NaOH totally ionizes, the pOH = 1. Since pH+pOH=14, pH=13.