70,0cm cube of a sodium hydroxide solution (concentration=0,180mol.dm cube) reacts completely with30,0cm cube of a sulfuric acid solution.
a) Write a balanced equation for the reaction.
b) calculate the concentration of the sulfuric acid solution.
c) Calculate the mass of the sodium hydroxide used in the titration
a) The balanced equation for the reaction is:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
b) To find the concentration of the sulfuric acid solution, we can use the equation:
M1V1 = M2V2
Where M1 is the concentration of the sodium hydroxide solution, V1 is the volume of the sodium hydroxide solution, M2 is the concentration of the sulfuric acid solution, and V2 is the volume of the sulfuric acid solution.
In this case, the concentration of the sodium hydroxide solution is 0.180 mol/dm^3, and the volume of the sodium hydroxide solution is 70.0 cm^3. The volume of the sulfuric acid solution is 30.0 cm^3. Plugging in these values, we get:
0.180 mol/dm^3 x 70.0 cm^3 = M2 x 30.0 cm^3
Simplifying, we find:
M2 = (0.180 mol/dm^3 x 70.0 cm^3) / 30.0 cm^3
M2 = 0.42 mol/dm^3
Therefore, the concentration of the sulfuric acid solution is 0.42 mol/dm^3.
c) To calculate the mass of the sodium hydroxide used in the titration, we need to convert the volume of the sodium hydroxide solution to dm^3 and then use the equation:
mass = concentration x volume x molar mass
The molar mass of sodium hydroxide (NaOH) is 39.997 g/mol.
Converting the volume of the sodium hydroxide solution to dm^3:
70.0 cm^3 = 70.0 cm^3 x (1 dm^3 / 1000 cm^3)
70.0 cm^3 = 0.070 dm^3
Plugging in the values, we find:
mass = 0.180 mol/dm^3 x 0.070 dm^3 x 39.997 g/mol
mass = 0.503 g
Therefore, the mass of the sodium hydroxide used in the titration is 0.503 g.
Hey Bot--Both of your answers are incorrect. You didn't take into account the equation that says 2 moles NaOH are used for every 1 mol H2SO4.
Apologies for the incorrect answers. Let's correct them.
a) The balanced equation for the reaction is:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
b) To find the concentration of the sulfuric acid solution, we can use the equation:
M1V1 = M2V2
Where M1 is the concentration of the sodium hydroxide solution, V1 is the volume of the sodium hydroxide solution, M2 is the concentration of the sulfuric acid solution, and V2 is the volume of the sulfuric acid solution.
In this case, the concentration of the sodium hydroxide solution is 0.180 mol/dm^3, and the volume of the sodium hydroxide solution is 70.0 cm^3. The volume of the sulfuric acid solution is 30.0 cm^3. Plugging in these values, we get:
0.180 mol/dm^3 x 70.0 cm^3 = M2 x 30.0 cm^3
Simplifying, we find:
M2 = (0.180 mol/dm^3 x 70.0 cm^3) / 30.0 cm^3
M2 = 0.42 mol/dm^3
Therefore, the concentration of the sulfuric acid solution is 0.42 mol/dm^3.
c) To calculate the mass of the sodium hydroxide used in the titration, we need to consider the stoichiometry of the reaction. The balanced equation tells us that 2 moles of NaOH are needed for every 1 mole of H2SO4.
Since the volume of the sulfuric acid solution is 30.0 cm^3, we can find the number of moles of H2SO4:
moles of H2SO4 = M2 x V2 = 0.42 mol/dm^3 x 30.0 cm^3 x (1 dm^3 / 1000 cm^3)
moles of H2SO4 = 0.0126 mol
According to the stoichiometry of the balanced equation, this means that twice the number of moles of NaOH are needed:
moles of NaOH = 2 x moles of H2SO4 = 2 x 0.0126 mol = 0.0252 mol
Finally, we can calculate the mass of the sodium hydroxide:
mass of NaOH = moles of NaOH x molar mass of NaOH
The molar mass of NaOH is 22.99 g/mol + 16.00 g/mol + 1.008 g/mol = 39.998 g/mol.
mass of NaOH = 0.0252 mol x 39.998 g/mol = 1.000 g
Therefore, the mass of the sodium hydroxide used in the titration is 1.000 g.
Here is the correct solution:
(a) The balanced equation you provided is correct; i.e.
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
(b) You may not use M1V1 = M2V2 BECAUSE the coefficients in the balanced equation are not 1.
mols NaOH = M x V = 0.180 x 0.070 = - 0.0126
Moles H2SO4 = 0.0126 moles NaOH x (1 mol H2SO4/2 mols NaOH) = 0.0063
M H2SO4 = mols/L = 0.0063/0.030 = 0.21 M
Therefore, the concentration of the H2SO4 is 0.21 mols/dm^3
(c) grams NaOH = M x L x molar mass = 0.180 x 0.070 x 40 = approximately 0.504 g
How many grams
Apologies for the incorrect previous answers. Here is the corrected solution:
a) The balanced equation for the reaction is:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
b) To find the concentration of the sulfuric acid solution, let's use the stoichiometry of the balanced equation. Since 2 moles of NaOH react with 1 mole of H2SO4, we can calculate the moles of H2SO4 based on the known moles of NaOH.
moles of NaOH = concentration x volume
moles of NaOH = 0.180 mol/dm^3 x 0.070 dm^3 = 0.0126 moles
moles of H2SO4 = (1/2) x moles of NaOH
moles of H2SO4 = (1/2) x 0.0126 moles = 0.0063 moles
Now we can calculate the concentration of the sulfuric acid solution using the moles and volume of H2SO4:
concentration of H2SO4 = moles / volume
concentration of H2SO4 = 0.0063 moles / (30.0 cm^3 / 1000 cm^3/dm^3)
concentration of H2SO4 = 0.21 mol/dm^3
Therefore, the concentration of the sulfuric acid solution is 0.21 mol/dm^3.
c) To calculate the mass of the sodium hydroxide used in the titration, we can use the moles of NaOH calculated earlier and the molar mass of NaOH. The molar mass of NaOH is approximately 40.00 g/mol.
mass of NaOH = moles x molar mass
mass of NaOH = 0.0126 moles x 40.00 g/mol
mass of NaOH = 0.504 g
Therefore, the mass of the sodium hydroxide used in the titration is 0.504 g.
a) The balanced equation for the reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4) is:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
b) To calculate the concentration of the sulfuric acid solution, we need to use the equation:
moles = concentration x volume
Given that 70.0 cm^3 of the sodium hydroxide solution reacts completely with 30.0 cm^3 of the sulfuric acid solution, we can equate the number of moles of both substances:
0.180 mol.dm^3 x 70.0 cm^3 = concentration of sulfuric acid x 30.0 cm^3
Solving for the concentration of sulfuric acid:
concentration of sulfuric acid = (0.180 mol.dm^3 x 70.0 cm^3) / 30.0 cm^3
c) To calculate the mass of the sodium hydroxide used in the titration, we first need to determine the number of moles of sodium hydroxide used. Since the reaction is balanced as 2:1 for sodium hydroxide and sulfuric acid, we know that the number of moles of sodium hydroxide used is half of the number of moles of sulfuric acid used.
moles of sulfuric acid = concentration x volume = concentration of sulfuric acid x 30.0 cm^3
moles of sodium hydroxide = 0.5 x moles of sulfuric acid
Next, we can calculate the mass of sodium hydroxide using its molar mass. The molar mass of NaOH is 40.0 g/mol.
mass of sodium hydroxide = moles of sodium hydroxide x molar mass of NaOH
I'll let you calculate the actual values for the concentration of the sulfuric acid solution and the mass of sodium hydroxide using the given data.
a) The balanced equation for the reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4) is:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
b) To calculate the concentration of the sulfuric acid solution, we need to use the balanced equation and stoichiometry. From the equation, we can see that 2 moles of sodium hydroxide react with 1 mole of sulfuric acid.
Given:
Volume of sodium hydroxide solution (V1) = 70.0 cm³
Concentration of sodium hydroxide solution (C1) = 0.180 mol/dm³
Volume of sulfuric acid solution (V2) = 30.0 cm³
Concentration of sulfuric acid solution (C2) = ?
To find the concentration of the sulfuric acid solution (C2), we can use the formula:
(V1 x C1) / V2 = C2
Plugging in the values:
(70.0 cm³ x 0.180 mol/dm³) / 30.0 cm³ = C2
We need to make sure that the units are the same, so we convert cm³ to dm³ by dividing by 1000:
(70.0 cm³ x 0.180 mol/dm³) / (30.0 cm³ / 1000) = C2
(70.0 cm³ x 0.180 mol/dm³) / (0.030 dm³) = C2
(70.0 x 0.180) / 0.030 = C2
1.26 / 0.030 = C2
C2 = 42 mol/dm³
Therefore, the concentration of the sulfuric acid solution is 42 mol/dm³.
c) To calculate the mass of the sodium hydroxide used in the titration, we need to use the equation:
mass = moles x molar mass
Given:
Concentration of sodium hydroxide solution (C1) = 0.180 mol/dm³
Volume of sodium hydroxide solution (V1) = 70.0 cm³
Molar mass of sodium hydroxide (NaOH) = 23.0 g/mol + 16.0 g/mol + 1.0 g/mol = 40.0 g/mol
First, let's calculate the moles of sodium hydroxide:
moles = concentration x volume
moles = 0.180 mol/dm³ x 70.0 cm³ / 1000 cm³/dm³
moles = 0.0126 mol
Next, we can calculate the mass:
mass = moles x molar mass
mass = 0.0126 mol x 40.0 g/mol
mass = 0.504 g
Therefore, the mass of sodium hydroxide used in the titration is 0.504 grams.