Answer to Question #178649 in General Chemistry for Brayden Stevens

Question

Titration:

Determine the molarity of a KOH solution when 14.7 mL of 0.180 M H₃PO₄neutralizes 25.0 mL of the KOH solution.

Steps:

1. Use the balanced neutralization reaction from Question 1. Read Section 10-7 in the textbook, and especially the paragraph called "Balancing Acid-Base Neutralization Equations".

2. Determine the moles of H₃PO₄ used to neutralize the reaction by using the volume and molarity of the H₃PO₄.  

3. Use the stoichiometry of the balanced reaction to fine the moles of KOH.

4. Divide the moles of KOH by the volume of the KOH solution, in L.

Answer 1

H₃PO₄+ 3KOH = K₃PO₄ + 3H₂O  

Number of moles of H₃PO₄ used for neutralization = (Molarity of H₃PO₄) x (Volume of H₃PO₄ in litres) = $0.180×[14.7×10^{−3}]=(2.646×10^{−3})moles$

Moles of KOH = $(2.646×10^{-3})×3=7.938×10^{-3} moles$

KOH solution = $\frac{7.938×10^{-3}}{25×10^{-3}}=3.1752×10^{-7} m$