For example:

What is the molarity of a $\ce{NaOH}$ solution if $25\rm~mL$ of the solution is exactly neutralized by $44\rm~mL$ of $0.32\rm~M~HCl$?

I know that the abstract function is

$$\text{Given Volume of Known Molarity} \times \text{Molarity Factor} \times \text{Mole Ratio} \times \text{Per Volume Factor} \times \text{Conversion Factor Per Liter}$$


closed as off-topic by Martin - マーチン Jul 2 '16 at 11:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ @ A.J what are your workings out so far? The policy on this site is to not straight out answer homework questions without going through the OP's workings out first $\endgroup$ – Technetium Feb 11 '16 at 7:21

The reaction that occurs: H+ + OH- --> H2O

When you multiply the volume of HCl by the molarity of HCl you get the total amount of moles of H+ that reacted (mol/L x L = mol). This means that the same amount of OH- was needed for the reaction (1:1). Volume of NaOH is known and thus yields the answer.

  • $\begingroup$ This doesn't answer the question or really make sense. $\endgroup$ – Technetium Feb 11 '16 at 7:19
  • $\begingroup$ @Joel I disagree. It does in a way answer the question and it makes perfect sense. However, it would have been better suited as a comment since it is neither complete nor does the question itself comply with our homework policy. $\endgroup$ – Martin - マーチン Jul 2 '16 at 11:29
  • $\begingroup$ @Martin , then why disagree? It's not the same amount , it's the same molar ratio. From there you can calculate the amount. $\endgroup$ – Technetium Jul 13 '16 at 4:01

Not the answer you're looking for? Browse other questions tagged or ask your own question.