25cm^3 of a solution of NaOH required 28cm^3 of 1.0 mol dm^-3 H2SO4 to neutralise it.

a)Write an equation for the reaction.

b)How many moles of H2SO4 were needed?

c)How many moles of NaOH were thus neutralised?

d)How many moles of NaOH are there in 25cm^3 of solution?

25cm^3 of a solution of NaOH required 28cm^3 of 1.0 mol dm^-3 H2SO4 to neutralise it.

a)Write an equation for the reaction.
2NaOH + H2SO4 ==> Na2SO4 + 2H2O

b)How many moles of H2SO4 were needed?
mols H2SO4 = M x L = ?

c)How many moles of NaOH were thus neutralised?
?mol H2SO4 x (2 mol NaOH/1 mol H2SO4) = ?mol H2SO4 x 2/1 = x


d)How many moles of NaOH are there in 25cm^3 of solution?
Isn't this the same answer as part c?

a) The equation for the reaction between NaOH and H2SO4 is:

2NaOH + H2SO4 -> Na2SO4 + 2H2O

b) To find the number of moles of H2SO4 needed, we use the equation:

Moles = Volume (in dm^3) x Concentration (in mol/dm^3)

Given that the volume of H2SO4 used is 28 cm^3 and the concentration is 1.0 mol/dm^3, we convert the volume to dm^3:

Volume = 28 cm^3 = 28/1000 dm^3 = 0.028 dm^3

Now we can calculate the moles of H2SO4:

Moles = 0.028 dm^3 x 1.0 mol/dm^3 = 0.028 mol

Therefore, 0.028 moles of H2SO4 were needed.

c) According to the balanced equation, the molar ratio between H2SO4 and NaOH is 1:2. Therefore, for every mole of H2SO4 used, 2 moles of NaOH are neutralized.

So, the number of moles of NaOH neutralized is 2 times the number of moles of H2SO4 used:

Moles of NaOH = 2 x 0.028 mol = 0.056 mol

Thus, 0.056 moles of NaOH were neutralized.

d) To find the number of moles of NaOH in 25 cm^3 of solution, we have to assume that the concentration of the NaOH solution is the same as the H2SO4 solution.

Since 1 dm^3 is equal to 1000 cm^3, we need to convert the volume of the solution to dm^3:

Volume = 25 cm^3 = 25/1000 dm^3 = 0.025 dm^3

Now we can calculate the number of moles of NaOH in the solution:

Moles = 0.025 dm^3 x 1.0 mol/dm^3 = 0.025 mol

Therefore, there are 0.025 moles of NaOH in 25 cm^3 of the solution.