In a standardization of the NaOH solution, you found that a beaker with 25.0 mL
ID: 512929 • Letter: I
Question
In a standardization of the NaOH solution, you found that a beaker with 25.0 mL of water with 5.55 g of KHP dissolved required 17.25 mL of NaOH to reach the equivalence point.
-How many moles of NaOH are in the solution?
-What is the actual concentration of the original NaOH solution?
-If an extra 25.0 mL of water were added to the NaOH solution before the titration with KHP, would more, less or the same amount of KHP have to be added to reach the equivalence point? (The molar mass of KHP is 204.22 g/mol)
Explanation / Answer
(a)
KHP + NaOH ----------> NaKP + H2O
Moles of KHP = mass / molar mass = 5.55 / 204.22 = 0.0272 mol
According to the above equation, 1 mol KHP = 1 mol NaOH
the, 0.0272 mol KHP = 0.0272 mol NaOH
(b) Molarity of NaOH = Moles of NaOH / Voulme of Solution in L = 0.0272 / 0.01725 = 1.58 M
(c) Addition of water does not vary the number of moles of NaOH. It only changes the concentration of solution. Hence, same number of moles of KHP would be needed.