Final answer:
To calculate the number of grams of KOH needed to neutralize 12.1 mL of 0.14 M HCl, you can use the balanced equation for the neutralization reaction between KOH and HCl and the formula moles = concentration x volume to find the number of moles of HCl. Since the molar ratio is 1:1, you will need the same number of moles of KOH. Then, you can use the molar mass of KOH to find the grams of KOH needed.
Step-by-step explanation:
To calculate the number of grams of KOH needed to neutralize 12.1 mL of 0.14 M HCl, we need to use the balanced equation for the neutralization reaction between KOH and HCl: KOH + HCl -> KCl + H2O. From the equation, we can see that the molar ratio between KOH and HCl is 1:1. This means that 1 mole of KOH reacts with 1 mole of HCl. To find the number of moles of HCl in 12.1 mL of 0.14 M HCl, we use the formula: moles = concentration x volume. Plugging in the values, we get: moles = 0.14 M x 0.0121 L = 0.001694 moles. Since the molar ratio is 1:1, we need the same number of moles of KOH.
Now, to find the grams of KOH, we need to use its molar mass. The molar mass of KOH is: 39.10 g/mol (K) + 16.00 g/mol (O) + 1.01 g/mol (H) = 56.11 g/mol. To find the grams of KOH, we multiply the number of moles (0.001694 moles) by the molar mass (56.11 g/mol): grams = 0.001694 moles x 56.11 g/mol = 0.09537 grams. Therefore, approximately 0.09537 grams of KOH are needed to neutralize 12.1 mL of 0.14 M HCl.