Final answer:
To neutralize 10.8 mL of 0.15 M HCl, calculate the moles of HCl and use the 1:1 molar ratio with KOH to determine the mass needed; 0.09089 grams of KOH are required.
Step-by-step explanation:
To calculate the amount of KOH needed to neutralize 10.8 mL of 0.15 M HCl, we need to use the stoichiometry of the neutralization reaction. The balanced chemical equation is:
HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l).
The reaction shows a 1:1 molar ratio between HCl and KOH. First, calculate the moles of HCl present:
Moles HCl = Volume (L) × Molarity (mol/L) = 0.0108 L × 0.15 mol/L = 0.00162 mol.
Since the ratio is 1:1, the moles of KOH needed are also 0.00162 mol. To find the mass of KOH required, multiply by the molar mass of KOH (56.11 g/mol):
Mass KOH = Moles KOH × Molar Mass (g/mol) = 0.00162 mol × 56.11 g/mol = 0.09089 g.
Therefore, 0.09089 grams of KOH are needed to neutralize 10.8 mL of 0.15 M HCl in stomach acid.