Question

A solution has a pH of 2.5. Its OH^- is 3.2 times 10^-12

How is this true?

Answer 1

@Photon336
@pooja195

Answer 2

@Whitemonsterbunny17

Answer 3

We must use this relationship.

pH + pOH = 14

If we know that the pH = 2.5

the pOH = 11.5

remember p = take the negative log -log of the concentration.

if

[OH] = 3.2x10^-12

take the -log[3.2x10^-12] and tell me what you get

Answer 4

@desmarie

Answer 5

I got 11.49485002, which I am assuming rounds up to 11.5.

Answer 6

@N8KD_Da_Beast

Answer 7

こんにちは(Kon'nichiwa) I'm here!! XD. How can I help u today?

Answer 8

the question listed above

Answer 9

Well looking at the work u and photon done, I can't help but agree with u.

Answer 10

but it is because photon is not on and I dont know what to do next

Answer 11

originally in the question it just gave me the pH. I got the answer to the OH^- by guessing but I want to know how to get the answer correctly

Answer 12

This questions remaining steps have me troubled as well. ごめんなさい(Gomen'nasai) but it seems that I can't be any help at the moment. さようなら(Sayōnara)

Answer 13

@Abhisar

Answer 14

Ok, so pH is basically -log[H+]. Right?

Answer 15

You have been given the concentration of OH- ions. So in order to calculate the pH you first need to calculate the concentration of H+ ions.

Answer 16

In any solution [H+] x [OH-] = \(\sf 10^{-14} \\ \Rightarrow [H+]=\frac{10^{-14}}{3.2 \times 10^{-12}}=3.12 \times 10^{-3}\)

Answer 17

Now, pH = -log\((3.12 \times 10^{-3})=2.5\)

Answer 18

Got it?

Answer 19

thank you so much

Answer 20

Welcome c:

Answer 21

sorry another question.
what if it gives me only the pH?

Answer 22

@Abhisar