Ok.I have a titration experiment yesterday 0.2 M of KHP were used and the volume of the titrated solution is 20Ml,it was titrated with 0.2M of NaOH solution with an equivalence volume of 18M.What is the theoretical Equivalence volume of KHP titrated
KHP + NaOH => KNaP + H2O
You took 20 mL of 0.2 M KHP and titrated that with 0.2 M NaOH. I don't have a clue as to the meaning of "equivalence volume of 18 M". Since both KHP and NaOH have the same molarity while KHP and NaOH contain one H and one OH, then the volume at the equivalence point should be 20 mL. (Perhaps you meant 18 mL and you just made a typo and omitted the L of mL).
To calculate the theoretical equivalence volume of KHP (potassium hydrogen phthalate) titrated, we need to use the balanced chemical equation for the reaction between KHP and NaOH.
The balanced equation is:
KHP + NaOH -> KNaP + H2O
From the equation, we can see that the mole ratio between KHP and NaOH is 1:1. This means that for every mole of KHP, we need 1 mole of NaOH to react completely.
Given that the volume of the titrated NaOH solution is 20 mL and its concentration is 0.2 M, we can calculate the number of moles of NaOH used:
Moles of NaOH = Concentration * Volume
Moles of NaOH = 0.2 M * 0.020 L
Moles of NaOH = 0.004 moles
Since the mole ratio between KHP and NaOH is 1:1, the number of moles of KHP titrated is also 0.004 moles.
Now, we can calculate the theoretical equivalence volume of KHP titrated by dividing the number of moles of KHP by the concentration of KHP:
Equivalence volume of KHP = Moles of KHP / Concentration of KHP
Given that the concentration of KHP used is 0.2 M, we can substitute the values:
Equivalence volume of KHP = 0.004 moles / 0.2 M
Equivalence volume of KHP = 0.02 L or 20 mL
Therefore, the theoretical equivalence volume of KHP titrated in this experiment is 20 mL.