Final answer:
To neutralize 20 ml of 0.5 M NaOH, 2.0422 grams of KHP (potassium hydrogen phthalate) is required, based on the 1:1 molar ratio of NaOH to KHP in the reaction.
Step-by-step explanation:
To calculate how much KHP (potassium hydrogen phthalate) in grams is needed to neutralize 20 ml of 0.5 M NaOH, we will use the concept of stoichiometry. The balanced chemical equation for the reaction between NaOH and KHP is:
NaOH + KHC8H4O4 → NaKC8H4O4 + H2O
This reaction is in a 1:1 molar ratio, meaning one mole of NaOH will neutralize one mole of KHP. First, we find the moles of NaOH:
Moles of NaOH = Volume (L) × Molarity (M) = 0.020 L × 0.5 M = 0.01 moles
Since the molar ratio of NaOH to KHP is 1:1, you'll also need 0.01 moles of KHP to fully neutralize the NaOH.
Molar mass of KHP (KHC8H4O4) = 204.22 g/mol
Moles of KHP = 0.01 moles
Mass of KHP = Moles of KHP × Molar mass of KHP = 0.01 moles × 204.22 g/mol = 2.0422 grams.
So, 2.0422 grams of KHP is needed to neutralize 20 ml of 0.5 M NaOH.