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If it takes 22.75 mL of 1.75 M H3PO4 to neutralize 14.90 mL of Al(OH)3, what is the molarity of the Al(OH)3?

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Molaritybase*volumebase*Heq=Molarityacid*volumeacid*Heq

Molaritybase*14.9ml*3=1.75*22.75ml*3

molaritybase= you do it...
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