### Calculating pH

Posted:

**Thu Dec 06, 2018 1:14 pm**How do you calculate the pH of an NaOH solution when given the molarity of the NaOH?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=60&t=38621

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Posted: **Thu Dec 06, 2018 1:14 pm**

How do you calculate the pH of an NaOH solution when given the molarity of the NaOH?

Posted: **Thu Dec 06, 2018 1:43 pm**

You will need to know the molarity of the NaOH. Let's assume the solution is 0.1M. NaOH is a strong base, so this will produce 0.1mol/L of OH ions in solution. This will produce a pH of 13.

You will need to take the negative log of 0.1 to find the pOH. This will work out to be 1.

Since pH + pOH = 14

We can calculate the pH to be 13.

This assumption can only be used for strong bases which dissociate completely in water.

You will need to take the negative log of 0.1 to find the pOH. This will work out to be 1.

Since pH + pOH = 14

We can calculate the pH to be 13.

This assumption can only be used for strong bases which dissociate completely in water.

Posted: **Thu Dec 06, 2018 1:50 pm**

For example, we are given that the molarity of a solution is 0.1M NaOH. Since we know that NaOH is a strong base, we know that it will fully dissociate. Therefore,

0.1 M NaOH --> 0.1 M OH-

pH = 14 - pOH = 14 - (-log([OH-])) = 14 - (-log([0.1M)) = 14 - 1 = 13

So, the final answer is pH = 13.

0.1 M NaOH --> 0.1 M OH-

pH = 14 - pOH = 14 - (-log([OH-])) = 14 - (-log([0.1M)) = 14 - 1 = 13

So, the final answer is pH = 13.