8 mL of 2.5 M phosphoric acid is neutralized with 500 mL of potassium hydroxide. What is the concentration of the base?
H3PO4 + 3KOH ==> K3PO4 + 3H2O
mols H3PO4 = M x L = 2.5 x 0.008 = ?
mols KOH = 3x that.
M KOH = moles KOH/L KOH.
To find the concentration of the base (potassium hydroxide), we can use the concept of stoichiometry and the balanced chemical equation of the neutralization reaction between phosphoric acid (H₃PO₄) and potassium hydroxide (KOH). The balanced equation for this reaction is:
H₃PO₄ + 3 KOH → K₃PO₄ + 3 H₂O
From the balanced equation, we can see that one mole of phosphoric acid reacts with three moles of potassium hydroxide. Thus, the moles of phosphoric acid can be determined by multiplying the volume (in liters) of the acid by its concentration (in moles per liter):
moles of H₃PO₄ = volume of H₃PO₄ (in liters) × concentration of H₃PO₄ (in moles per liter)
Substituting the given values into the equation:
moles of H₃PO₄ = 8 mL ÷ 1000 (to convert to liters) × 2.5 M = 0.02 moles
Since the reaction is 1:3 (mole ratio) between H₃PO₄ and KOH, the moles of potassium hydroxide can be determined:
moles of KOH = 3 × moles of H₃PO₄ = 3 × 0.02 = 0.06 moles
Now, we can calculate the concentration of the base (KOH) by dividing the moles of KOH by the volume of the base:
concentration of KOH = moles of KOH ÷ volume of KOH (in liters)
However, we need to convert the volume of the base (given in milliliters) to liters:
volume of KOH = 500 mL ÷ 1000 (to convert to liters) = 0.5 L
Finally, substituting the values into the equation:
concentration of KOH = 0.06 moles ÷ 0.5 L = 0.12 M
Therefore, the concentration of the base (potassium hydroxide) is 0.12 M.