In a standardization of the NaOH solution, you found that a beaker with 25.0 mL
ID: 967314 • 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
KHC8H4O4 + NaOH ç=====èNaKC8H4O4 + H2O
KHP = Potassium Hydrogen phthalate
KHP Molecular weight: 204.22 g/mol
Amount of KHP used: 5.55 g
No. of moles of KHP = 5.55/204.22 = 0.0272 Moles
Volume of water for dissolving KHP = 25 mL = 0.025L………….V1
Molarity of KHP = 0.0272/0.025 = 1.088 M………………………..M1
Volume of NaOH used = 17.25 mL=0.01725L…………………….V2
Since M1V1 = M2V2
Therefore Molarity of NaOH, M2 = M1V1/V2 = 1.088 x 0.025 / 0.01725 = 1.577 M
No. of moles of NaOH = 1.577 x 0.01725 = 0.0272 Moles
If additional 25 mL (0.025 L) of water is added, total volume of NaOH solution = 0.025L + 0.01725L = 0.04225 L
Molarity of original NaOH solution = 1.577M
Original volume of NaOH solution = 0.01725 L
Total volume of dilute NaOH solution = 0.04225L
Molarity of diluted NaOH solution = 1.577 x 0.01725 / 0.04225 = 0.644M
No. of moles of NaOH in diluted solution = 0.644 x 0.04225 = 0.0272 Moles
Since the number of moles of NaOH has not changed before and after dilution of NaOH solution, it’s expected that
same amount of KHP solution is required for reaching equivalence point in titration.