How do you calculate the molarity of NaOH? I am just very confused on which formula I use to calculate.

I am given the equation for calculating the concentration of NaOH with (molarity of acid)(volume of acid)(#hydrogen ions)in acid =(molarity of base )(volume of base)(# of hydroxide ions) in base.

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Well, you have left out the increments we need to answer the question. But I think I can help guide you in the right direction, so here goes!!!

The first thing to understand about molarity is it is the number of moles of solute (NaOH) per liter of solvent (H2O, or water). To calculate the number of moles of NaOH calculate the molar mass of NaOH. Use the periodic table of the elements, and you will find sodium has an atomic mass of 23, oxygen has an atomic mass of 16, and hydrogen has an atomic mass of 1. Add those all together, and you get 40 grams per mole, for one mole of NaOH.

Now, if we knew the amount of NaOH used to make the solution...I will just go ahead and use 20 grams of mass, measured out on an electronic balance. You take the 40 grams/mole we just calculated above, divide that by the 20 grams of NaOH we just measured on the scale. Notice the grams cancel out and we are left with 2 moles of NaOH.

Lets pretend we used 500 milliliters of water for the solvent, into which we dissolved our NaOH. That would be .5 liters of water. Remember molarity is moles of solute per liter of water, so take the 2 moles of NaOH and divide that by the .5 liter of water, and we get a NaOH solution with a molarity of 4.

There is another excellent example included on the resource link I provided for you. Hope this helps!

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