chemistry questions about titration.........?

Question

1) 21. 23.45 mL of 0.275 M sodium hydroxide was used to titrate against mL of acetic acid. What was the concentration in M of acetic acid?

2) 24. 24.92 mL of 0.00199 M silver nitrate was used to titrate against 5 mL of sodium chloride solution. What was the concentration of NaCl?

3) 23. 35.79 mL of 0.275 M sodium hydroxide was used to titrate against 15 mL of sulfuric acid. What was the concentration in M of sulfuric acid?

please explain how you did it, thanks.

Answer 1

1) can't do it,vol. not known.
2)I suppose this means 24.92 ml of 0.00192 is reacted with 5ml of sodium nitrate. on that basis this is how it's done
Vol(1) x Molity(1) = Vol(2) x molity(2) so to find molity (2)
rearrange to molity(2) = Vol(1)xmolity(1)/vol2,put in numbers
molarity of sodium chloride=24.92x1.99/5 x 1000
0.00991816
To find concentration of NaCl in g/l remember that one molar NaCl contains 23+35.5 g/l = 58.55 g/l
so your test solution contains0.00991816 x 58.55 g/l
=0.0.580708268 g/l=0.581g/l(3 sf)

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You need to practice this or you will never be confident; so you do the others.Just make sure how many molecules of A react wih how many of B.So Write out the equation to check(Hint in Question 3 notice 2 moles of NaOH react with 1 mole of H2SO4)

Good luck,nose to the grindstone,So long as you get a few correct you'll be OK

Answer 2

Use the equation :-

Moles(n) = [Conc(mol dm^-3)] x vol(cm^3) / 1000(cm^3)

NB
[ 1000 cm^3 to bring everything to cm^3.


Q1). Cannot do because you have not quoted the volume of acetic acid (mL)

Q2). AgNO3 + NaCl = NaNO3(aq) + AgCl(s)
Molar ratios are all one - to - one.
n(AgNO3) = 0.00199 mol Dm^-3 x 24.92 mL / 1000 mL
n(AgNO3) = 4.959 x 10^-5 moles
n(NaCl) = 4.959 x 10^-5 moles (Because ratios are 1:1)
By algebraic rearrangement :-
[Conc] = moles x 1000 / vol
[NaCl] = 4.959 x 10^-5 x 1000 / 5 = 9.918 x 10^-3 moles dm^-3
or Conc'n (NaCl) = 9.918 x 10^-3 M = 0.009918M

3).
2NaOH + H2SO4 = Na2SO4(aq) + 2H2O(l)
Molar ratios are 2:1::1:2
n(NaOH) = 0..275 x 35.79 / 1000 = 9.842 x 10^-3 moles
n(H2SO4) = 9.842 x 10^-3 x 0..5 = 4.921 x 10^-3 moles (Because ratios are 2:1)
By algebraic rearrangement :-
[Conc] = moles x 1000 / vol
[H2SO4] = 4.921 x 10^-3 x 1000 / 15 = 3.281 x 10^-1 moles dm^-3
or Conc'n (H2SO4) = 3.281 x 10^-1 M = 0.3281M

Hope this helps!!!!