20 mL of NaOH needed to titrate 30 mL if 6 M HCl, what is the molarity of NaOH?

I totally forget how to solve titraton problems! help please?

Answer:
Ok. First, notice we have the titration of a strong acid (HCl) with a strong base (NaOH). It's important to know this because not all titrations result in all of the reactants being converted to products
.
Now the best way to think of the probem is to recal that you need to balance this equation. It is, after all, simpy a reaction where you have a certain quantity of reactants and products.

20mL of NaOH with an unknown concentration, X.
We first want to see how many moles of HCl we have. Recall that 1M concentration is 1 mole/Liter. Here we have 30mL of 6 M
30mlL= .03L
We multiply through to obtain moles:
.03L* (6moles/1L)= .18 moles of HCl
Normally we have to balance out, but notice that 1 mol HCl combines with 1 mol NaOH to give H20 and NaCl. Therefore, to completely titrate .18 moles HCl, we need .18 moles of NaOH.

All we do now is solve for the unknown X.
20mL= .02L* X= .18 moles
When we solve for X, we get .18/.02=9M

Therefore, a 9M concentration of 20mL of NaOH is required to titrate 30mL of 6M HCL

The rule of thumb is Concentration(Base)*Volume=
Concentration(Acid)*Volume.

There's a nice shortcut here in which you don't have to convert to Liters, because as long as both units are in mL the equation stays the same. In fact, check for yourself:

20(whatever units, say mL)*9=30 (mL, keep units same)*6

180=180
As long as you keep track of units, that will give you a fast solution.

Caveat: This formula applies to strong acid/base reactions. Weak acid/base require dissociation constants and are more tricky. Also, keep track of the (H+) and (OH-) and make sure you balance them out. H2SO4 (Sulphuric Acid) gives 2H+ per mole, not just 1. That is, keep track of Normality, which is the Molarity * number of H+ or OH- that dissociate in reaction.
.030 L x 6M HCl = ___ moles HCl

Since 1 mole HCl = 1 mole NaOH,

___ moles NaOH / .020 L NaOH = ___ M NaOH

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