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Chem Help...clutch if someone figures this out
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The number of mL of .0144 M HCL needed to completely neutralize 420mg of KOH is _____
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It's been a while since I took chemistry but I believe the answer is 1042mL. Here's my reasoning...
.420g(1 mol / 28 g)(6.022*10^23 atoms/mol)(1L/((.0144mol)(6.022*10^23atoms/mol)))(1000mL/L)
I can't figure out a better way to write it, but you set the number of OH atoms your .420g sample equal to the number of H atoms in your other sample, and convert the answer to liters (or mL).
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I used to know how to do this like 3 years ago.
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By the way, it's pretty badass that you can use google as a calculator to solve all that shit.
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nvm got it...for those wondering:
Both strong acid(H)/base(OH) 1-1 to make water (neutralize)
.0144 moles of H+ in one liter
.0075 moles of OH- (420mg->.42 g-> .42/56 (KOH MM of 56)->.0075 m of KOH -> 1-1 ion ratio OH also .0075
.0074/.0144 *1000
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except that the molar mass of KOH is 28, not 56.
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