70,0cm^3 of a sodium hydroxide solution ( concentration= 0,180 mol•dm^-3 ) reacts completely with 30,0 cm^3 of a sulfuric acid solution. Calculate the mass of the sodium hydroxide used in the titration. ( 0,540 g)

Answer 1

To solve this problem, we need to use the concept of stoichiometry. The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4) is:

2NaOH + H2SO4 -> Na2SO4 + 2H2O

From the equation, we can see that 2 moles of NaOH react with 1 mole of H2SO4.

First, we need to calculate the number of moles of H2SO4 in the 30.0 cm^3 of solution. To do this, we need to convert cm^3 to dm^3:

30.0 cm^3 = 30.0 / 1000 = 0.030 dm^3

Next, we can calculate the number of moles of H2SO4:

moles of H2SO4 = concentration * volume
= 0.180 mol•dm^-3 * 0.030 dm^3
= 0.0054 mol

Since the stoichiometry of the reaction is 2:1 for NaOH to H2SO4, the number of moles of NaOH used is twice the number of moles of H2SO4:

moles of NaOH = 2 * 0.0054 mol
= 0.0108 mol

Finally, we can calculate the mass of NaOH:

mass of NaOH = moles of NaOH * molar mass of NaOH
= 0.0108 mol * 40.0 g/mol
= 0.432 g

Therefore, the mass of sodium hydroxide used in the titration is 0.432 g.

Answer 2

Bot, this is incorrect. None of the work with H2SO4 is needed.
moles NaOH = M x L = 0.180 x 0.070 = 0.0126
grams NaOH = moles NaOH x molar mass NaOH = 0.0126 mol x 40 g/mol = 0.0504 g NaOH

Answer 3

Apologies for the mistake.
Using the correct approach, the number of moles of NaOH can be calculated using the concentration of the NaOH solution (0.180 mol/dm^3) and the volume used (70.0 cm^3):

moles NaOH = concentration x volume
= 0.180 mol/dm^3 * 0.070 dm^3
= 0.0126 mol

Then, the mass of NaOH can be calculated by multiplying the number of moles by the molar mass of NaOH (40 g/mol):

mass NaOH = moles NaOH x molar mass NaOH
= 0.0126 mol x 40 g/mol
= 0.504 g NaOH

Therefore, the correct mass of sodium hydroxide used in the titration is 0.504 g.