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Kaelin J
Molarity of NaOH (Standardization)?
Need help trying to find the Molarity of NaOH.
Here's what I know.
Mass of dry NaOH: .683g
Mass of final solution: 169.213g
Percent by mass NaOH: .404%
Mass of dry KHP: 3.141g
Mass of KHP contained in each 25.00 mL sample: .3141g
Moles of KHP contained in each 25.00 mL sample: .001538 Moles KHP
Standardization of NaOH
Final V (mL) / Initial V (mL) / V NaOH (mL) / V NaOH (L) / Molartity of NaOH
24.4 / 0.00 / 21.6 / .0216 / Need Help
27.8 / 0.00 / 22.2 / .0222 / Need Help
27.7 / 0.00 / 22.3 / .0223 / Need Help
27.6 / 0.00 / 22.4 / .0224 / Need Help
27.9 / 0.00 / 22.1 / .0221 / Need Help
The 0's in the table above represent the burets initial volume, it starts at 0 and works it's way down to 50.00ml, so since it says 0.00mL, that just means that's what it was started out, the amount we used is the V NaOH (third column). The fifth column is what I'm confused about, I can't finish the rest of my report until I figure out how I do this.
How we thought we were supposed to find the Molarity of NaOH was Moles over liter since that is the formula. We tried this by doing .683g NaOH (dry mass of NaOH) divided by 40.00g NaOH (Molar Mass) to give us the moles of NaOH we used. Then do that over the liters we used, (column 4). Example of this would be .0171Moles / .0284 L = .602 M, apparently this answer is 5 or so times larger than what we should've gotten.
Any help would be much appreciated. Sorry about the long explanation, just wanted to put you up to speed and maybe help us out for where we went wrong. THANKS!
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