Calculate the pH of solutions having the following ion concentrations at 298 K.

Question

a) [H⁺] = 2.0 ✕ 10^-2 M

(b) [H⁺] = 2.0 ✕ 10^-6 M

(c) [OH^-] = 7.3 ✕ 10^-6 M

Answer 1

Katie T.,

That's at least 3 questions in the same general area at the same time, not an efficient use of Expert answerers' time or of your learning capacity. Please select one problem, accept the hints given, finish the solution and summarize to yourself the required reasoning and procedures. Then submit another problem, only if new reasoning is required. Otherwise you have an excellent chance of copying the solution, without increasing your understanding at all.?

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Answer 2

pH is defined as the negative log of the hydronium ion concentration, or

pH = -log [H3O+] = -log [H+]

a) [H⁺] = 2.0x10^-2 M

pH = -log 2.0x10^-2 M

pH = 1.70

b) [H⁺] = 2.0x10^-6 M

pH = -log 2.0x10^-6 M

pH = 5.70

c) [OH^-] = 7.3x10^-6 M

To find [H⁺] you can use [H⁺][OH^-] = 1x10^-14 and then...

[H⁺] = 1x10^-14 / 7.3x10^-6 = 1.37x10^-9

pH = -log 1.37x10^-9

pH = 8.86

or you can find pOH which is -log [OH^-]

pOH = -log 7.3x10^-6 M

pOH = 5.14 and then use pH + pOH = 14 and solving for pH, we have..

pH = 14 - 5.14

pH = 8.86