# Homework Help: Calculating pH

1. Dec 19, 2011

### Saitama

1. The problem statement, all variables and given/known data
x mole of a strong monoacidic base is dissolved in one litre of water. The pH of the solution will be
(a)-log x
(b)14-log x
(c)14+log x
(d)-log (14-x)

2. Relevant equations

3. The attempt at a solution

For instance i assumed that the base is NaOH but it does not react with water to form any compound. So i am confused how should i start?

Thanks!

2. Dec 19, 2011

### chemisttree

I'm still working on "monoacidic base"...
What's that?

OK. I just Googled it. So you have a base that generates 1 mole of OH- per mole of base in water. You also know the following:
1) pH = -log[H+]
2) [H+][OH-] = 1.01 X 10-14

Can you get it from there?

Last edited: Dec 19, 2011
3. Dec 19, 2011

### agent99

First calculate the pOH, which equals to -log[OH-].
Then, pH=14-pOH. That's it.

4. Dec 19, 2011

### Saitama

I think it isn't stated that it generates one mole of OH-.

5. Dec 23, 2011

### sankalpmittal

Monoacidic base is written in the form of MOH where M is any univalent metal.

Hints :
1 mole of MOH in water furnishes one OH- ion.
Then x moles of MOH will furnish.... ?

Write an ionic equation in terms of "x" and think about Mole Concept once again. You can then use the formulas given by chemisttree.

Last edited: Dec 23, 2011
6. Dec 23, 2011

### Staff: Mentor

Huh? What do the laws concerning gases have to do with the pH of the solution?

7. Dec 23, 2011

### Saitama

This isn't a hard question as i thought of and thanks to chemistree for the idea of pOH.

I found out the pOH which is -log x. Then calculated pH, which is completely easy pH=14-pH.

@sankalp:- I think i am not that bad at chemistry that you need to remind me of Mole concept.

I just forgot to reply in this thread after solving the problem. :P

8. Dec 23, 2011