Question

Hello , I'm struggling in the following questions and I need an explanation if it is possible

question 1) :A 51.0 mL aliquot of HCl(aq) of unknown concentration was titrated with 0.226 M NaOH(aq). It took 102.4 mL of the base to reach the endpoint of the titration. what the concentration (M) of the acid was ?

question 2) :How many moles of BaCl2 are formed in the neutralization of 196.5 mL of 0.095 M Ba(OH)2 with aqueous Hal?

question 3) : Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide: Pb2+(aq) + 2I-(aq) → PbI2(s)

Lead iodide is virtually insoluble in water so that the reaction appears to go to completion. How many milliliters of 1.180 M HI(aq) must be added to a solution containing 0.200 mol of Pb(NO3)2(aq) to completely precipitate the lead?

question 4) :What are the respective concentrations (M) of Fe3+ and I- afforded by dissolving 0.300 mol FeI3 in water and diluting it to 750 mL?

question 5 ) :What is the concentration (M) of a NaCl solution prepared by dissolving 7.6 g of NaCl in sufficient water to give 490 mL of solution?

Answer 1

Q. 1

Solution : HCl and NaOH neutralization reaction is,

HCl + NaOH ---> NaCl + H2O

So, 1 mole of HCl is neutralized with 1 mole of NaOH

moles of NaOH (base) = molarity x volume = 0.226 M x 0.1024 L = 0.023 mole

So moles of HCl (acid) = 0.023 mole

concentration (M) of acid = moles/L of solution = 0.023 mole / (0.051 + 0.1024) L = 0.15 M

Q. 2

Solution: the chemical equation would be,

Ba(OH)2 + 2HCl ----> BaCl2 + 2H2O

So, 1 mole of Ba(OH)2 reacts with 2 mole of HCl to give 1 mole of BaCl2

moles of Ba(OH)2 = molarity x volume = 0.095 M x 0.1965 = 0.019 mole

So moles of BaCl2 formed = 0.019 moles

Q 3.

Solution : As can be seen from the given chemical equation, 1 mole of Pb2+ require 2 moles of I- to form 1 mole of PbI2.

moles of Pb(NO3)2 = 0.200 mole

So, moles of HI = 2 x 0.200 = 0.400 moles

Volume of HI (aq) solution required to completely precipitate Pb2+ = moles/molarity = 0.4 mole/1.180 M = 0.339 L

= 33.90 ml

Q. 4

Solution : Molarity = moles/L

So, molarity of FeI3 = 0.300 mole/0.750 L = 0.4 M

every FeI3 has 1 Fe and 3 I

So concentrations of,

[Fe3+] = 0.4 M

[I-] = 3 x 0.4 = 1.2 M

Q. 5

Solution : Molarity = moles of solute/L of solution

moles of solute = grams of solute/molar mass of solute = 7.6 g / 58.44 g/mol = 0.13 mol

So, concentration (M) = 0.13 mol / 0.490 L = 0.26 M