Please explain how to do this question?:
What volume of 0.040mol/L H2CO3(aq) is required to neutralize 18.5mL of 0.15mol/L Ca(OH)2(aq)?
What volume of 0.040mol/L H2CO3(aq) is required to neutralize 18.5mL of 0.15mol/L Ca(OH)2(aq)?
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Neutralization occurs when the number of moles of acid are equated to those of the base. Multiply the volume of the hydroxide by its molarity to find the number of basic moles. Use this value to then find the volume of acid necessary to undergo neutralization.
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H2CO3 is Carbonic acid, a weak acid, and diprotic. It will neutralise Calcium Hydroxide....
H2CO3(aq) + Ca(OH)2(aq) --------> CaCO3(aq) + 2H2O(l)
Moles of Ca(OH)2(aq) = 0.15 x 0.0185 = 0.002775 moles
we need this much H2CO3(aq), so it must be 0.002775 / 0.040 = 0.069375L OR 69.4mL
H2CO3(aq) + Ca(OH)2(aq) --------> CaCO3(aq) + 2H2O(l)
Moles of Ca(OH)2(aq) = 0.15 x 0.0185 = 0.002775 moles
we need this much H2CO3(aq), so it must be 0.002775 / 0.040 = 0.069375L OR 69.4mL