50.0 mL of 0.200 M NaOH neutralized 20.0 mL of sulfuric acid. Determine the
concentration of the acid. Show a balanced chemical equation, all work, and the answer
rounded to three decimal places.
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
mols NaOH = M x L = approx 0.01
mols H2SO4 = 1/2 that from the coefficients in the balanced equation.
M H2SO4 = mols H2SO4/L H2SO4 = ?
0.250M
To determine the concentration of the sulfuric acid, we'll use the balanced chemical equation for the reaction:
2 NaOH + H2SO4 → Na2SO4 + 2 H2O
According to the equation, it takes 2 moles of NaOH to react with 1 mole of H2SO4. Therefore, the moles of H2SO4 can be calculated using the following equation:
moles of H2SO4 = (moles of NaOH) / 2
Let's calculate the moles of NaOH first:
moles of NaOH = volume (in L) × concentration (in mol/L)
= 50.0 mL × (1 L/1000 mL) × 0.200 mol/L
= 0.010 mol
Now we can calculate the moles of H2SO4:
moles of H2SO4 = (0.010 mol NaOH) / 2
= 0.005 mol
Next, we need to determine the concentration of the sulfuric acid. The concentration is given by the equation:
concentration (in mol/L) = moles / volume (in L)
We already have the moles of H2SO4 as 0.005 mol. To find the volume, we sum the volumes of NaOH and H2SO4 used in the reaction. The volume of NaOH is 50.0 mL and the volume of H2SO4 is 20.0 mL.
total volume = volume of NaOH + volume of H2SO4
= 50.0 mL + 20.0 mL
= 70.0 mL
Now, we convert the volume to L:
total volume = 70.0 mL × (1 L/1000 mL)
= 0.070 L
Finally, we can calculate the concentration of the sulfuric acid:
concentration = moles / volume
= 0.005 mol / 0.070 L
≈ 0.071 mol/L
Therefore, the concentration of the sulfuric acid is approximately 0.071 mol/L, rounded to three decimal places.
To determine the concentration of the sulfuric acid, we first need to write a balanced chemical equation for the neutralization reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4):
2NaOH + H2SO4 -> Na2SO4 + 2H2O
From the equation, we can see that 2 moles of NaOH react with 1 mole of H2SO4 to produce 1 mole of Na2SO4 and 2 moles of water.
To find the amount of sulfuric acid in moles, we can use the concept of stoichiometry. Since the ratio between NaOH and H2SO4 is 2:1, we know that the moles of NaOH used would be half of the moles of H2SO4.
Given that the volume of NaOH solution is 50.0 mL = 0.0500 L and the concentration of NaOH is 0.200 M, we can calculate the moles of NaOH used:
moles of NaOH = volume (L) x concentration (M)
moles of NaOH = 0.0500 L x 0.200 M = 0.0100 moles
Since the ratio between NaOH and H2SO4 is 2:1, the moles of H2SO4 neutralized would be half of the moles of NaOH:
moles of H2SO4 = 0.0100 moles / 2 = 0.0050 moles
Given that the volume of sulfuric acid solution is 20.0 mL = 0.0200 L, we can now calculate the concentration of the acid:
concentration of H2SO4 = moles / volume (L)
concentration of H2SO4 = 0.0050 moles / 0.0200 L = 0.250 M
Therefore, the concentration of the sulfuric acid is 0.250 M (rounded to three decimal places).