Trial 1 0.379 0.30 18.51 18.21 0.00184 0.00184 0.100 Trial 2 Trial 3 Trial 4 0.3

Question

Trial 1 0.379 0.30 18.51 18.21 0.00184 0.00184 0.100 Trial 2 Trial 3 Trial 4 0.321 0.273 0.293 1.Mass of KHP used (g) 2.Initial Buret Reading (mL) 3. Final Buret Reading (mL) [4] Volume of NaOH solution added (mL) [5] Moles KHP used (mol) [6] Moles of NaOH reacted with KHP (mol) [7] Concentration of NaOH sol. (mol/L) [8] Average NaOH concentration (mol/L) [9] Standard Deviation (can use-STDEVS formula) [10] Percent Relative Standard Deviation (Q.1) Using your calculated concentration of NaOH from lab how many grams of KHP would you need to start with 0.50 19.10 18.51 0.00157 0.00157 0.0848 1.29 19.11 17.82 0.0133 0.0133 0.73 20.20 19.47 0.00143 0.00143 0.0734 0.0178 0.0690 0.0358 51.90% in order to have an end point at 35.00 mL? (Q.2) If 0.839 g of KHP were titrated, and 34.58 mL NaOH solution were necessary to reach the equivalence point what is the concentration of the NaOH solution? (Make sure to report the correct number of significant figures)

Explanation / Answer

Q1

mass of KHP needed to start in order to get an endpoint of V = 35 mL

concentration of NaOH --> 0.069 M

Volume of NaOH --> (18.21+18.51+17.82+19.47)/4 = 18.5 mL

then...

mmol of NaOH = MV = 18.5*0.069 = 1.2765 mmol of NaOH

we need

1.2765 mmol of KHP

mass = mmol*MW = 1.2765*204.22 = 260.6 mg of KHP

Q2.

m =0.839 g of KGP

V = 34.58 mL of NaOH required...

find NaOH concentration..

mol of KHP = 0.839 /204.22 = 0.00410 mol

mol of NAOH = 0.00410

[NaOH] = mol/V = 0.00410 / (34.58*10^-3) = 0.1185 M

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.