1. What is the molarity of a solution of H2SO4 is 42.7cm3 is required to exactly titrate 27.5cm3 of 0.61M NaOHhere is what i got:

H2SO4 + 2NaOH --> Na2SO4 + 2H2O(l)Mole ratio of H2SO4 to NaOH = 1:2therefore moles NaOH = 0.61 x 27.5= 16.775 moltherefore moles of H2SO4 needed = 16.775/2= 8.3875 molesconc = mol/volM Na2SO4 = 8.3875/42.7=0.196M

The answer is correct (should really be 0.20 M, to 2 sig fig, as only 2 sig fig in qestion).HOWEVER, your working is incorrect, as you have made two mistakes that cancel one another out. In determining the number of moles of NaOH, you have multiplied concentration, in mol/L, by volume in mL. As a result, you have an incorrect calculation, as n(NaOH) = 0.016775 mol. You should either include a 10^(-3) factor to correct volume into L, ie:

therefore moles NaOH = 0.61 x 27.5 x 10^(-3)= 16.775 x 10^(-3) mol

or you should work in millimoles (mmol), ie:

therefore moles NaOH = 0.61 x 27.5= 16.775 mmol

The reason that you have the correct answer is that in the last step, you have divided by volume, again in mL, and so the two 10^(-3) factors that you have missed would have cancelled.

quote:

2. How many cm3 of 0.48M H2SO4 is required to neutralise 26.0cm3 of 0.61M NaOH?

H2SO4 + 2NaOH „³ Na2SO4 + 2H2O(l)Mole ratio of H2SO4 to NaOH is 1:2 concentration = moles/volumeconc. x vol (H2SO4) = 2[conc. x vol. (NaOH)]0.48v = 2(0.61 x 26)vol(H2SO4) = 2(0.61 x 26)/0.48= 66.1 cm3

This one is wrong, the correct answer is 17 cm3 (to 2 sig fig). The mistake lies in the statement that"conc. x vol (H2SO4) = 2[conc. x vol. (NaOH)]".

You have said that n(H2SO4) : n(NaOH) = 1 : 2Upon cross multiplying, this gives 2 * n(H2SO4) = n(NaOH)and thus your statement should read 2 * conc. x vol (H2SO4) = [conc. x vol. (NaOH)].

You are also working in mL instead of L again.

quote:

3. To what volume must 400cm3 of 0.61M NaOH be diluted to yield a 0.50 solution

conc = mol/vol0.61= mols/400= 244 mol0.5 = 244/volvol = 488cm3

This is the correct answer, although you again have used mL instead of L, and thus have chemical amount in mmol instead of mol. Note that this question could be done quicker using the dilution formula.

I have posted some approaches to mole calculations in another recent thread, if you want to go have a look.

Hope this helps.

QuickQuote

quote:

Calculate the mass of solid required of 250 mL of 0.100 mol/L sodium carbonate solution

I interpret this to mean calculate the mass of sodium carbonate solid required to prepare 250 mL of 0.100 mol/L solution...

n(Na2CO3) = CV = 0.100 * 250 *10^(-3) = 0.025 mol

Now, assuming we are using anhydrous sodium carbonate (Na2CO3), M = 105.99 g/mol, thenm(Na2CO3) = nM = 0.025 * 105.99 = 2.64975 = 2.65 g (3 sig fig)

If we use hydrated sodium carbonate, Na2CO3.10H2O, M = 286.15 g/mol, thenm(Na2CO3.10H2O) = nM = 0.025 * 286.15 = 7.15375 = 7.15 g (3 sig fig)

As for the question using oxalic acid dihydrate, I'll leave you to try it, but the answer I get is 0.946 g (3 sig fig).

Oh, and I'm glad you like my method for mole calculations.

I have another question:0.3162 g of oxylaic acid (H2C2O4.2H2O) was dissolved in water and the solution made up to 250.0 mL. Calculate the concentration of the solution. Do I include the 2H2O in the mole calculation??? I can never remember and then I get confused and its a big mess of messiness

You must include the water of crystallisation in the calculation of the moles, as it is present as part of the mass. That is, the 0.3162 g of oxalic acid dihydrate is the mass of the oxalic acid present, and the mass of the water of crysatllisation present in the crystal lattice. Thus,

n(H2C2O4) = n(H2C2O4.2H2O) = m / M = 0.3162 / (2 * 1.008 + 2 * 12.01 + 4 * 16.00 + 2 * 2 * 1.008 + 2 * 16.00)= whatever (I don't have a calculator on hand, sorry)

So, [H2C2O4] = n / V = whatever / (250.0 * 10^(-3)) = answer mol/L (4 sig fig)

Originally posted by Elliot IIII've only been taught the "dodgy method" for titration. Could someone please explain to me the other method? Thanks!

Step 1: Write a balanced equation for the reaction taking place.

Step 2: Below each formula, write down the information that you have - use this to identify the substance that you WANT to find out about, and the substance that you've GOT enough information to figure out moles.

Step 3: Figure out moles of GOT, usually from a formula like n = m / M, or n = C * V

Step 4: Write down n(WANT) / n(GOT) = coeff(WANT) / coeff(GOT), but using the formulae for WANT and GOT on LHS, and the values from you balanced equation on the RHS

Step 5: Find n(WANT) from n(GOT) using the equation in step 4.

Step 6: Answer the question by calculating whatever you seek from n(WANT) using the same sorts of equations as used in step 3.

NOTE: This method will allow you to solve all mole calculations, not just the ones involved with titrations.

To make this clearer, let's try an example...

25.00 ml of an unknown potassium hydroxide solution was placed into a clean, dry conical flask, and 2 drops of a suitable indicator added. The solution was then titrated with 0.2371 mol/L sulfuric acid, and 22.35 mL of solution was required to reach the end point. Determine the concentration of the potassium hydroxide solution.

Equation: 2KOH (aq) + H2SO4 (aq) ---> K2SO4 (aq) + 2H2O (l)Data: 25.00 mL 22.35 mLC = ? 0.2371 mol/LSo... WANT GOT

moles(GOT): Now, n(H2SO4) = CV = 0.2371 * 22.35 * 10^(-3) = 5.2991... * 10^(-3) mol

Eqn: FROM EQUATION n(KOH) / n(H2SO4) = 2 / 1

moles(WANT): So, n(KOH) = (2 / 1) * n(H2SO4) = 2 * 5.2991... * 10^(-3) = 0.010598... mol

Answer: And so, [KOH] = n / V = 0.010598... / (25.00 * 10^(-3)) = 0.42393... mol/L

So, the concentration of the unknown KOH solution is 0.4239 mol/L (4 sig fig).

Originally posted by kalindawhy dissolve the stuff straight into a beaker and not just into the volumetric flask

The problem with this is if the solid does not readily dissolve, there's not much you can do about it in the volumetric flask. For example, you can't heat it, as that would change the internal volume of the vol flask, and ruin it.

Substances like iron(II) sulfate, for example, readily dissolve at high temperatures, but not at low temperatures - this despite being highly soluble. That is, once you get them dissolved, they stay dissolved, but getting them to dissolve can be a challenge.

quote:

Originally posted by kalindathis way there is least chance of having any of the solution left in the beaker

True, and this is a potential source of human error, but correct quantitative transfer of a solution should avoid this problem. So, one stuff in beaker dissolved, transfer (through a filter funnel) into the vol. flask. Then, rinse the beaker 3 times, each time placing the solution into the vol. flask. Then, rinse the filter funnel 3 times. By doing this, all of the chemical is transferred to the vol. flask.

Calculate the mass of solid required of 250 mL of 0.100 mol/L sodium carbonate solution

I interpret this to mean calculate the mass of sodium carbonate solid required to prepare 250 mL of 0.100 mol/L solution...

n(Na2CO3) = CV = 0.100 * 250 *10^(-3) = 0.025 mol

Now, assuming we are using anhydrous sodium carbonate (Na2CO3), M = 105.99 g/mol, thenm(Na2CO3) = nM = 0.025 * 105.99 = 2.64975 = 2.65 g (3 sig fig)

If we use hydrated sodium carbonate, Na2CO3.10H2O, M = 286.15 g/mol, thenm(Na2CO3.10H2O) = nM = 0.025 * 286.15 = 7.15375 = 7.15 g (3 sig fig)

As for the question using oxalic acid dihydrate, I'll leave you to try it, but the answer I get is 0.946 g (3 sig fig).

Oh, and I'm glad you like my method for mole calculations.

C6H8O7+3H2 < -------- > C6H5O73- + 3H3O+1

Thats for complete ionisation.

For the intermediates you can work out yourself.

That should be 3H2O at the start, not 3H2 .

Citric acid (2-hydroxypropane-1,2,3-tricarboxylic acid), HOOC-CH2-C(OH)(COOH)-CH2-COOH is a triprotic acid, and like all triprotic acids, it ionises in a stepwise manner. ie:

C6H8O7 (aq) + H2O (l) <---> C6H7O7- (aq) + H3O+ (aq)

C6H7O7- (aq) + H2O (l) <---> C6H6O72- (aq) + H3O+ (aq)

C6H6O72- (aq) + H2O (l) <---> C6H5O73- (aq) + H3O+ (aq)

All three equilibria lie to the left, each further left than the one before - so, not many citric acidmolecules undergo the first ioinisation, and virtually none the third. In fact, many questions treat it as if it were a monoprotic acid - which is wrong, but you should recognise that it can happen, especially in assessments and trials, which may be written by people with not as much chemistry knowledge as would be desirable.