Help with Chemistry?

Question

how would you calculate the pH of 0.5M NaOH, or 2 M HCl, or 3e-8 M HCl?

Answer 1

If you know the concentration of the H+ or OH- ions in solution you can use this information to find the pH of the solution (or the pOH, from which you can calculate pH).

You are given two strong electrolytes in the question.

Sodium Hydroxide, NaOH, is a strong base. When you dissolve NaOH in water it will completely disassociate into Na+ and OH- ions. For every 1 mole of NaOH dissolved you will obtain 1 mole of Na+ and 1 mole of OH- ions. You are given the concentration of NaOH in solution and by knowing the mole ratio of how many moles of NaOH dissolved yields how many moles of OH- ions in solution (1:1), we also know the concentration of OH- ions.

HCl is a strong acid. HCl will completely disassociate into H+ and Cl- ions in solution. For every 1 mole of HCl dissolved in solution you will get 1 mole of H+ ions and 1 mole of Cl- ions. You are given the concentrations of HCl in solution so (using the same process as with the NaOH), you can find the concentrations of H+ ions.

** We have assumed here that,
[NaOH] = [OH-] and
[HCl] = [H+]
This assumption works at “high” concentrations of acids and bases where the only significant donor of H+ or OH- ions is the acid/base itself. But when the concentration drops down to be on the order of water’s own contribution…we must take that into account too. Even a very dilute solution of an acid/base on the order of E-5 Molar is virtually unaffected by the disassociation of water into H+ and OH- ions since the acid is more than 100 times more concentrated than the water’s contribution. But in the last part of your question, you ate given a VERY dilute HCl concentration that our assumption will not yield the correct answer.
Water will naturally disassociate into H+ and OH- ions. The concentration of both of these ions at equilibrium is 1 E-7 Molar. If you were to calculate the pH due to these ions, you would get a pH of 7 (neutral).
When you find the pH on the last part of the problem (with the [HCl] = 3 E-8 M, you need to add the water’s [H+] to the HCl’s [H+] to get the total [H+] to find the pH, otherwise you will get an alkaline answer…which is obviously wrong.


pH = -log [H+]
pOH = -log [OH-]

pH + pOH = 14

Where using brackets, [ ], refers to the concentration of the substance inside the brackets in terms of Molarity.