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wrote...
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A month ago
An aliquot( 28.7 mL ) of a KOH solution required 31.3 mL of 0.118 M HCI for neutralization. What mass ( g ) of KOH was in the original sample?

A)0.129
B)0.190
C)0.173
D)0.207
E)0.215
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wrote...
Educator
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A month ago Edited: A month ago, bio_man
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> 0.207

The reaction is: . . .KOH + HCl ==> H2O + NaCl

moles HCl = M HCl x L HCl = (0.118)(0.0313) = 0.00369 moles HCl

The balanced equation tells us that KOH and HCl react in a 1:1 mole ratio.

0.00369 moles HCl x (1 mole KOH / 1 mole HCl) = 0.00369 moles KOH

0.00369 moles KOH x (56.1 g NaOH / 1 mole NaOH) = 0.207 g KOH
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wrote...
Educator
A month ago
Topic updated Upwards Arrow
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wrote...
A month ago
Thank you so much, but from where do u get the 56.1 g.
wrote...
Educator
A month ago
That's the molar mass of NaOH
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wrote...
A month ago
Ohh ok ok, thanks again
wrote...
Educator
A month ago
Ohh ok ok, thanks again

You're welcome. Let us know if you have any other questions (chem, math, etc)
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wrote...
A month ago
Hello I was calculating the molar mass and that not the answer that I got. I get 39.99g
wrote...
Educator
A month ago
Sorry, I meant molar mass of KOH 🤦‍♂️
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wrote...
A month ago
Ohh ok it's fine thank you so muchh☺️
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