Answer :
Answer:
a. H₂CO₃(aq) + KOH(aq) ⇄ K₂CO₃(aq) + H₂O(l)
b. 0.603 M
Explanation:
Step 1: Write the neutralization reaction
H₂CO₃(aq) + KOH(aq) ⇄ K₂CO₃(aq) + H₂O(l)
Step 2: Calculate the reacting moles of KOH
22.9 mL of 1.430 M KOH react.
0.0229 L × (1.430 mol/L) = 0.0327 mol
Step 3: Calculate the reacting moles of H₂CO₃
The molar ratio of H₂CO₃ to KOH is 1:1. The reacting moles of H₂CO₃ are 1/1 × 0.0327 mol = 0.0327 mol.
Step 4: Calculate the molarity of H₂CO₃
0.0327 moles of H₂CO₃ are in a volume of 54.2 mL. The molarity of H₂CO₃ is:
M = 0.0327 mol/0.0542 L = 0.603 M