Self Test Check

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FDeCastro_1B
Posts: 54
Joined: Thu Jul 25, 2019 12:16 am

Self Test Check

Postby FDeCastro_1B » Thu Dec 05, 2019 2:28 pm

Just checking to see if I have the right answer since I can't find it online.

Calculate the pH of 0.077 m NaOH(aq).

ShravanPatel2B
Posts: 100
Joined: Fri Aug 30, 2019 12:18 am

Re: Self Test Check

Postby ShravanPatel2B » Thu Dec 05, 2019 2:40 pm

pOH = -log(.077) = 1.11
pH = 14-1.11 = 12.9

hope this helps!

FDeCastro_1B
Posts: 54
Joined: Thu Jul 25, 2019 12:16 am

Re: Self Test Check

Postby FDeCastro_1B » Thu Dec 05, 2019 2:41 pm

ShravanPatel4G wrote:pOH = -log(.077) = 1.11
pH = 14-1.11 = 12.9

hope this helps!


Yes! Thank you.


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