Question

Good morning! I've been struggling to answer some of my practice problems. Thank you for your help.

1. Strontium chloride and sodium fluoride react to form strontium fluoride and sodium chloride, according to the reaction shown.

SrCl2(aq)+2NaF(aq)⟶SrF2(s)+2NaCl(aq)SrCl2(aq)+2NaF(aq)⟶SrF2(s)+2NaCl(aq)

What volume of a 0.530 M NaF0.530 M NaF solution is required to react completely with 969 mL969 mL of a 0.300 M SrCl20.300 M SrCl2 solution? volume:____ mL?

How many moles of SrF2SrF2 are formed from this reaction?

moles of SrF2SrF2: ___ mol?

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

Pb(NO3)2(aq)+2NH4I(aq)⟶PbI2(s)+2NH4NO3(aq)Pb(NO3)2(aq)+2NH4I(aq)⟶PbI2(s)+2NH4NO3(aq)

What volume of a 0.1700.170 M NH₄I solution is required to react with 801801 mL of a 0.4000.400 M Pb(NO₃)₂ solution?

volume: ____ mL?

How many moles of PbI₂ are formed from this reaction? moles: ____ mol PbI2?

3.The balanced equation for the neutralization reaction of aqueous H2SO4H2SO4 with aqueous KOHKOH is shown.

H2SO4(aq)+2KOH(aq)⟶2H2O(l)+K2SO4(aq)H2SO4(aq)+2KOH(aq)⟶2H2O(l)+K2SO4(aq)

What volume of 0.2700.270 M KOHKOH is needed to react completely with 19.019.0 mL of 0.2400.240 M H2SO4H2SO4? volume: ___ mL?

4. Consider the neutralization reaction

2HNO3(aq)+Ba(OH)2(aq)⟶2H2O(l)+Ba(NO3)2(aq)2HNO3(aq)+Ba(OH)2(aq)⟶2H2O(l)+Ba(NO3)2(aq)

A 0.1150.115-L sample of an unknown HNO₃ solution required 35.935.9 mL of 0.1500.150 M Ba(OH)₂ for complete neutralization. What is the concentration of the HNO₃ solution? concentration: ___ M?

5. Assuming equal concentrations and complete dissociation, arrange these aqueous solutions by their freezing points. Highest freezing point to the Lowest freezing point. Answer bank: Na2CO3, Li3PO4, KNO3.

Answer 1

Q3:

millimole of KOH used = 2× millimole of sulphuric acid

0.270M× V(ml) = 2× 19.0ml × 0.240 M

Volume of KOH used = 33.78ml

Q4:

Millimole of HNO₃ = 2× millimole of Ba(OH)₂

0.115×1000ml × M1 = 2 × 0.150M × 35.9ml

M1 = 0.0937 M

Q5:

Depression in freezing point is directly proportional to the number of ions of solute particle. I = 4 for lithium phosphate, I=3 for sodium carbonate and I=2 for potassium nitrate.

Order is

(Highest freezing point) KNO₃ >Na₂CO₃ > Li₃PO₄ (lowest freezing point)