Answer:
To calculate the mass of KHP (potassium hydrogen phthalate) needed to use 35.00 mL of 0.400 M NaOH in a titration, we first need to determine the balanced chemical equation for the reaction between KHP and NaOH.
KHP reacts with NaOH in a 1:1 ratio to form potassium sodium phthalate (KNaC8H4O4) and water:
KHP + NaOH → KNaC8H4O4 + H2O
We can use this balanced equation to determine the moles of NaOH needed to react with 1 mole of KHP.
From the equation, we can see that 1 mole of NaOH reacts with 1 mole of KHP. Therefore, the number of moles of NaOH required to react with a given amount of KHP can be calculated using the following equation:
moles of NaOH = moles of KHP
Next, we need to calculate the number of moles of NaOH that will react with 35.00 mL of 0.400 M NaOH:
moles of NaOH = volume x concentration
moles of NaOH = 0.03500 L x 0.400 mol/L
moles of NaOH = 0.014 mol
Since we need the same number of moles of KHP, we can use this value to calculate the mass of KHP needed:
moles of KHP = 0.014 mol
molar mass of KHP = 204.22 g/mol
mass of KHP = moles of KHP x molar mass of KHP
mass of KHP = 0.014 mol x 204.22 g/mol
mass of KHP = 2.86 g
Therefore, you would need 2.86 g of KHP to use 35.00 mL of 0.400 M NaOH in a titration.