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hlilly2413
 one year ago
Hey, I can't figure out what to do with this problem. It states "
in an attempt to titrate a 50.0mL solution of 2.50M KOH, you accidentally added 57.4mL of 3.05M HCl, which you then realized was too much. To obtain a neutralized solution, you know you need to add a base, but all you have left is 1.15M NaOH. How much NaOH (in mL) should you add?"
So, I set up the original equation as 2KOH +2HCl > 2KCl +2H20 and figured out the # of moles and the grams of both of the reactants.
I know that the base should neutralize the acid, but I just don't know what to do next. Thanks!
hlilly2413
 one year ago
Hey, I can't figure out what to do with this problem. It states " in an attempt to titrate a 50.0mL solution of 2.50M KOH, you accidentally added 57.4mL of 3.05M HCl, which you then realized was too much. To obtain a neutralized solution, you know you need to add a base, but all you have left is 1.15M NaOH. How much NaOH (in mL) should you add?" So, I set up the original equation as 2KOH +2HCl > 2KCl +2H20 and figured out the # of moles and the grams of both of the reactants. I know that the base should neutralize the acid, but I just don't know what to do next. Thanks!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438275273827:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For get the L (volume) you need to: mol/M \[mol/(mol/L)= L\]
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