Final answer:
0.0821 grams of KOH are required to neutralize 12.2 mL of 0.12 M HCl in stomach acid, using the molar ratio of the reaction between HCl and KOH, which is 1:1.
Step-by-step explanation:
To determine how many grams of KOH are needed to neutralize 12.2 mL of 0.12 M HCl in stomach acid, we can use the concept of molarity and the stoichiometry of the neutralization reaction. HCl and KOH react in a 1:1 molar ratio according to the equation:
HCl + KOH → KCl + H2O
First, we calculate the moles of HCl:
- Moles HCl = volume (L) × concentration (M)
- Moles HCl = 0.0122 L × 0.12 mol/L = 0.001464 mol
Since the molar ratio is 1:1, we need the same number of moles of KOH to neutralize the HCl. Now, we calculate the mass of KOH needed:
- Molar mass of KOH = 39.1 g/mol (K) + 16.0 g/mol (O) + 1.0 g/mol (H) = 56.1 g/mol
- Mass of KOH = moles × molar mass
- Mass of KOH = 0.001464 mol × 56.1 g/mol = 0.0820984 g
Therefore, 0.0820984 grams (rounded to 0.0821 grams) of KOH are needed to neutralize 12.2 mL of 0.12 M HCl in stomach acid.