How can I find the pH of a solution prepared by the addition of 100mL of 0.15M HCl solution to 100mL of 0.2M of NaOH?
This wouldn't be as hard if I was given a Ka or Kb value, but I'm not.
Thanks
2007-04-01 12:49:56 · 2 answers · asked by bigmanonthewall 3 in Science & Mathematics ➔ Chemistry
I see where your going now. What would be the equation and table?
2007-04-01 13:09:55 · update #1
This is a strong acid / strong base neutralization which goes past the end point. You have 100 x 0.15 = 15 meq H+ mixed with 100 x 0.2 = 20 meq OH- thus there is a net of 5 meq of OH- dissolved in 200 ml. Concentration of OH is thus 5/200 = 0.025
pOH = - log 0.025 = 1.6, pH = 14 - 1.6 = 12.4
Ka's not involved when you have SA/SB.
2007-04-01 13:17:33 · answer #1 · answered by Robert J 2 · 0⤊ 0⤋
You don't need Ka or Kb because in low concentrations both HCl and NaOH are essentially completely ionized. Determine the concentration of the reactant that is left over after the full amount of NaCl is produced. The concentration of H+ or OH- in that will give you the pH
2007-04-01 19:55:34 · answer #2 · answered by gp4rts 7 · 1⤊ 0⤋