Question

The ionization constant for water (Kw) is 2.9 x 10^(-14) at 40 °C. Calculate [H3O+], [OH-], pH, and pOH for pure water at 40 °C.

Name: the ionization constant for water kw is 29 x 1014 at 40 c calculate h3o oh ph and poh for pure water at 40 c 93476
Uploaded: 2022-05-21T14:36:40-08:00
Duration: 3 min
Channel: Sima Sarker
Description: the ionization constant for water kw is 29 x 1014 at 40 c calculate h3o oh ph and poh for pure water at 40 c 93476

          The ionization constant for water (Kw) is 2.9 x 10^(-14) at 40 °C. Calculate [H3O+], [OH-], pH, and pOH for pure water at 40 °C.

Added by Timothy E.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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Solved by Expert Sima Sarker

05/21/2022

Step 1

Given: Kw = 2.9 x 10^(-14) [H3O+] = [OH-] = sqrt(Kw) [H3O+] = sqrt(2.9 x 10^(-14)) [H3O+] = 1.7 x 10^(-7) mol/L Show more…

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The ionization constant for water (Kw) is 2.9 x 10^(-14) at 40 °C. Calculate [H3O+], [OH-], pH, and pOH for pure water at 40 °C.

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The ionization constant for water (Kw) is 2.9 x 10^(-14) at 40 °C. Calculate [H3O+], [OH-], pH, and pOH for pure water at 40 °C.

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