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.
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)
3 answers
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
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
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.
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.