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# A2 Chemistry Titration question!!?Help!! Watch

1. I remember learning this topic about 2 months ago but don't understand this question (although it's simple).

**A 30.0 cm3 sample of a 0.480 mol dm–3 solution of potassium hydroxide was partially neutralised by the addition of 18.0 cm3 of a 0.350 mol dm–3 solution of sulphuric acid.

(i) Calculate the initial number of moles of potassium hydroxide.

0.0144 moles

(ii) Calculate the number of moles of sulphuric acid added.

0.0063 moles

This is the part I don't get(iii) Calculate the number of moles of potassium hydroxide remaining in excess in the solution formed.

Based on the equation H2sO4 + 2KOH -> K2SO4 + 2H20 I initially thought that the excess moles of KOH left over would've been 0.0144-0.0063 moles however my vague notes from class say that it is 0.0144-2(0.0063) moles

Is it that 2 moles of alkali have to neutralise 2 moles of acid.
Thanks.
2. hey, well you have H2SO4 + 2KOH and you have 0.0063 mol of the H2SO4 and 0.0144 mol of KOH.

However the molar ratio of sulphuric acid and potassium hydroxide is 1:2

Therefore, with 0.0063 mol of H2SO4, there should be twice the amount of mol of KOH that's reacted with it, which is 0.0124 (0.0063 x 2). So, we have 0.0124 mol of KOH thats reacted with all of the H2SO4, but you have more than 0.0124 mol, you have 0.0144 mol. The mol of KOH thats in 'leftover' is 0.0144 - 0.0124 which is 0.0081 mol. hope that explains it well
3. (Original post by tropicalia_la)
hey, well you have H2SO4 + 2KOH and you have 0.0063 mol of the H2SO4 and 0.0144 mol of KOH.

However the molar ratio of sulphuric acid and potassium hydroxide is 1:2

Therefore, with 0.0063 mol of H2SO4, there should be twice the amount of mol of KOH that's reacted with it, which is 0.0124 (0.0063 x 2). So, we have 0.0124 mol of KOH thats reacted with all of the H2SO4, but you have more than 0.0124 mol, you have 0.0144 mol. The mol of KOH thats in 'leftover' is 0.0144 - 0.0124 which is 0.0081 mol. hope that explains it well
Yes you've explained it really well, but I have really bad cognitive processing So I'm still a bit confused lol
4. I really hate chemistry
5. (Original post by lilzoldier)
I really hate chemistry
Haha
6. (Original post by tropicalia_la)
0.0124 (0.0063 x 2).
0.0144 - 0.0124 which is 0.0081 mol.
You have the right method, just your working out is wrong.
0.0063 x 2 = 0.0126
0.0144 - 0.0124 = 0.002

(Original post by Jin3011)
Yes you've explained it really well, but I have really bad cognitive processing So I'm still a bit confused lol
Basically because sulphuric acid contains 2 H+, for every mole of sulphuric acid, you need two moles of potassium hydroxide to neutralise it. This means the total amount of alkali needed to neutralise the acid is 0.0063x2 = 0.0126mol.

Thus any alkali added extra to this amount will not react (as all the sulphuric acid has been used up) and will remain in the solution. So 0.0144-0.0126 = 0.0018mol.
You have the right method, just your working out is wrong.
0.0063 x 2 = 0.0126
0.0144 - 0.0124 = 0.002

Basically because sulphuric acid contains 2 H+, for every mole of sulphuric acid, you need two moles of potassium hydroxide to neutralise it. This means the total amount of alkali needed to neutralise the acid is 0.0063x2 = 0.0126mol.

Thus any alkali added extra to this amount will not react (as all the sulphuric acid has been used up) and will remain in the solution. So 0.0144-0.0126 = 0.0018mol.
Hey thanks. I kind of figured it out yesterday when looking through an ancient chem equations book.
+rep for helping anyways =].
8. glad you managed to work it out!
You have the right method, just your working out is wrong.
0.0063 x 2 = 0.0126
0.0144 - 0.0124 = 0.002

Basically because sulphuric acid contains 2 H+, for every mole of sulphuric acid, you need two moles of potassium hydroxide to neutralise it. This means the total amount of alkali needed to neutralise the acid is 0.0063x2 = 0.0126mol.

Thus any alkali added extra to this amount will not react (as all the sulphuric acid has been used up) and will remain in the solution. So 0.0144-0.0126 = 0.0018mol.

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