Question

A student performs a neutralization reaction between potassium hydroxide and phosphoric acid. How many moles of KOH is required to neutralize 35.5 ml of 0.4599 M H₃PO₄ based on the given, unbalanced reaction below?

a) 0.045 moles
b) 0.091 moles
c) 0.183 moles
d) 0.227 moles

Answer 1

To neutralize 35.5 ml of 0.4599 M H₃PO₄, the moles of H₃PO₄ are first calculated and then multiplied by 3, according to the balanced chemical equation, yielding approximately 0.0489 moles of KOH needed. The closest given option to this value is (a) 0.045 moles.

Step-by-step explanation:

To determine how many moles of KOH are required to neutralize 35.5 ml of 0.4599 M H₃PO₄, we first use the balanced chemical equation for the neutralization reaction:

H₃PO₄ (aq) + 3KOH (aq) → K₃PO₄ (aq) + 3H₂O (l)

Note that the balanced equation requires three moles of KOH for every mole of H₃PO₄. Next, we calculate the moles of H₃PO₄ in 35.5 ml of the 0.4599 M solution:

moles of H₃PO₄ = volume (L) × concentration (M) = 0.0355 L × 0.4599 M = 0.0163 moles of H₃PO₄

Since the reaction requires 3 moles of KOH per mole of H₃PO₄, we multiply the moles of H₃PO₄ by 3:

moles of KOH = 3 × moles of H₃PO₄ = 3 × 0.0163 moles = 0.0489 moles

The closest option to 0.0489 moles of KOH is 0.045 moles. Therefore, the correct answer is (a) 0.045 moles.