Question

Recall that the equivalence point in a titration is the point when the amount of titrant added is just enough to react completely with the analyte solution. For this titration, phenolphthalein is selected as the indicator because the color change for the endpoint is near the equivalence point.

For 1.750g of KHP, how many milliers of 0.3 M sodium hydroxide (NaOH) is needed to reach the endpoint of the titration? Report your answer to the nearest milliliter. Choose the single best answer.
A. 35 ml
B. 12 ml
C. 82 mL
D. 29 ml

Answer 1

To reach the endpoint of the titration, we need to determine the amount of sodium hydroxide (NaOH) needed to react completely with the analyte solution, which in this case is 1.750g of potassium hydrogen phthalate (KHP). The volume of 0.3 M NaOH required to reach the endpoint is 29 mL (D).

Step-by-step explanation:

To reach the endpoint of the titration, we need to determine the amount of sodium hydroxide (NaOH) needed to react completely with the analyte solution, which in this case is 1.750g of potassium hydrogen phthalate (KHP). We can use the formula: moles of KHP = mass of KHP / molar mass of KHP. Then, we use the balanced equation to find the moles of NaOH required: moles of NaOH = moles of KHP. Finally, we use the molarity of NaOH to calculate the volume: volume of NaOH = moles of NaOH / molarity of NaOH.

First, we calculate the moles of KHP: moles of KHP = 1.750g / (204.22 g/mol) = 0.0085714 mol. This is the moles of NaOH required as well.

Now, we calculate the volume of NaOH: volume of NaOH = 0.0085714 mol / 0.3 M = 0.0285714 L = 28.57 mL.

Rounding to the nearest milliliter, the answer is 29 mL (D).

Answer 2

For 1.750g of KHP, 29 mL of 0.3 M NaOH is required to reach the endpoint of the titration, corresponding to option D. This is calculated using stoichiometry and the one-to-one mole ratio of KHP to NaOH, considering the molar mass of KHP is approximately 204.22 g/mol.

Step-by-step explanation:

To calculate the volume of 0.3 M sodium hydroxide (NaOH) needed to reach the endpoint of the titration with 1.750g of potassium hydrogen phthalate (KHP), we can use stoichiometry. KHP has a molar mass of roughly 204.22 g/mol, and the reaction between KHP and NaOH is a one-to-one mole ratio:

KHP (C₈H₅KO₄) + NaOH → NaKP (C₈H₄KO₄) + H₂O

First, we find the moles of KHP:

Moles of KHP = mass of KHP / molar mass of KHP = 1.750 g / 204.22 g/mol = 0.00857 mol

Since the mole ratio of KHP to NaOH is 1:1, we need 0.00857 mol of NaOH to reach the equivalence point. Next, we can calculate the volume of NaOH solution required:

Volume of NaOH = moles of NaOH / molarity of NaOH = 0.00857 mol / 0.3 mol/L

Volume of NaOH = 0.02857 L, which is 28.57 mL. Therefore, we round to the nearest milliliter, giving 29 mL as the answer.

Therefore, for 1.750g of KHP, 29 mL of 0.3 M NaOH is needed to reach the endpoint of the titration, corresponding to option D.