A titration neutralizes 25mL of an unknown acid using 32 mL of 6.0 M KOH. What is the molarity of the unknown acid in the original solution?

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A titration neutralizes 25mL of an unknown acid using 32 mL of 6.0 M KOH. What is the molarity of the unknown acid in the original solution?

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Answer 1 · 2017-04-30T00:31:20

$[(M o l a r i t y A c i d) (V o l u m e A c i d)] = [(M o l a r i t y B a s e) (V o l u m e B a s e)]$ $[(Molarity Acid)(Volume Acid)] = [(Molarity Base)(Volume Base)]$

Explanation:

$(M o l a r i t y A c i d) (25 m l) = (6.0 M) (32 m l)$ $(Molarity Acid)(25 ml) = (6.0M)(32ml)$

$M o l a r i t y A c i d = \frac{(6.0 M) (32 m l)}{(25 m l)} = 7.68 M$ $Molarity Acid = [(6.0M)(32ml)]/[(25ml)] = 7.68M$

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A titration neutralizes 25mL of an unknown acid using 32 mL of 6.0 M KOH. What is the molarity of the unknown acid in the original solution?

1 Answer

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