Thread: Stoichiometry

can someone help me with this question?(btw its just miscellaneous exercises in my textbook)

When HCl is added to aqueous lead(II) nitrate, solid lead(II) chloride is precipitated. If 10cm^3 of 2mol dm^-3 HCl is added to 40cm^3 of 0.5 mol dm^-3 aqueous lead nitrate, determine the concentration in the final solution of;

a)nitrate ions
b)chloride ions
c)hydrogen ions
d)lead(II) ions

thank you

Does your book give the answer? Is this correct?

a) 0.8 moles/dm-3
b) 0.0 moles/dm-3
c) 0.8 moles/dm-3
d) 0.2 moles/dm-3

Thank you for trying to help me, but im really more interested in the workings, rather than the answer. This darn book doesn't give any answers to thousands of questions in this book.......

I haven't done this for years (decades actually) and thought I'd tell you the answers I got first, so someone could correct me if I got it wrong. Carrying on regardless, here's how I worked it:

Calculate the moles of each reagent.

10 cm3 of 2 mol/dm3 of HCl contains 2 x 10/1000 moles = 0.02 moles HCl
40 cm3 of 0.5 mol/dm3 of Pb(NO3)2 contains 0.5 x 40/1000 = 0.02 moles lead nitrate.

Since the equation for the reaction shows you need 2 moles of HCl for each mole of Pb(NO3)2 you don't have enough HCl to react with all the lead nitrate, so you can see right away there will be no Cl- ions left in the solution; hence b) must be zero.

You can also see that half of the lead nitrate will be unreacted so there will be 0.01 moles of Pb-2 left in solution. The total volume of solution is 50 cm3 so the answer to d) is 0.01 /50 * 1000 = 0.2 mols/dm-3.

All of the nitrate ions remain in solution so you have 0.02 x 2 moles of NO3- in 50 dm-3, which is .02 x 2/50 x 1000 so the answer to a) is 0.8 mols/dm-3.

By similar rasoning all of the H+ ions remain in solution so you again have 0.02 x 2 (since you needed 2 HCL molecules per lead nitrate molecule to balance the equation) x 50/1000 = 0.8 mols/dm-3 which is the answer to c).

thank you for that, i think it makes sense.

i need urgent help with this topic, so can you try helping me solve one more?? i have attempted at home, but im not getting there.

1.552g of pure carboxylic acid (Y-COOH) is titrated against 0.4822 mol dm^-3 aqueous sodium hydroxide and 26.35cm^3 are found to be required for complete neutralisation. calculate the molar mass o the acid and hence deduce the probable formula.

thank you.

I had to look up the reactions to find that the terminal H is replaced by an Na. This means there is one mole of NaOH for one mole of Y-COOH. Calculate the moles of NaOH from the volume and concentration of the titrant. This gives you the moles of NaOH and of the acid since they must be equal. Since you also have the mass of acid you can easily calculate its molar mass.

From there I would probably use trial and error, trying different combinations of H and C for Y until I came to one that gave the correct molar mass.