swartzlanderar swartzlanderar
  • 12-04-2021
  • Chemistry
contestada

Given [OH-] 6.98 times 10^-2 what is the ph and [H+]

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MrScienceGuy
MrScienceGuy MrScienceGuy
  • 12-04-2021

Answer:

[H+]=1.43*10^-13

pH=12.84

Explanation:

Take the -log of the [OH-] to get the pOH:

pOH= -log(6.98*10^-2)=1.15

pH=14-pOH

pH=14-1.15

pH=12.843

[H+]=10^(-pH)

[H+]=10^(-12.843)

[H+]-1.43*10^-13

If I helped, a brainliest would be greatly appreciated!

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