1. A flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator.

Question

1. A flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. A buret contains 0.190 M NaOH.

A) What volume of NaOH is needed to reach the end point of the titration?

B) What was the initial concentration of HCl?

2. Lead (II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate.

A. What volume of 0.130 M NH4I solution is required to react with 321 mL of a 0.260 M Pb(NO3)2 solution?

B. How many moles of PbI2 are formed from this reaction?

3. What is the balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI? Include phases.

A. What mass of precipitate will form if 1.50 L of highly concentrated Pb(ClO3)2 is mixed with 0.500 L of 0.140 M NaI? Assume the reaction goes to completion.

Explanation / Answer

1.

Your question contains 2 unknowns -1) the concentration of the HCl , and the volume of NaOH solution used. It is impossible to calculate both unknowns. You must provide the data for 1 of these. Then the second unknown can be calculated.

2.

Pb(NO3)2 + 2 NH4I ? PbI2 + 2 NH4NO3

a) (0.321 L) x (0.2000 M) x (2/1) / (0.130 M) = 0.987 L NH4I

b) (0.321 L) x (0.2000 mol/L) x (1/1) = 0.642 mol PbI2

3.

Pb(ClO3)2 (aq) & 2 NaI (aq) --> PbI2 (s) & 2 NaClO3 (aq)


find moles
0.50 L (0.140 mol / Litre) NaI = 0.07 moles of NaI

by the equation
Pb(ClO3)2 (aq) & 2 NaI (aq) --> PbI2 (s) & 2 NaClO3 (aq)
0.07 moles of NaI produces 1/2 as many moles of PbI2 = 0 035 moles PbI2

use molar mass to find grams:
0 035 moles PbI2 ( 461.02 grams / mol PbI2) = 13.1 grams of PbI2

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

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