During a titration experiment, a 150.0 mL solution of 0.05 M sulfuric acid (H2SO4) is neutralized by 300.0 mL solution with an unknown concentration of sodium hydroxide (NaOH).
What is the concentration of the sodium hydroxide solution?
The answer I got is, 0.25 M
the acid is diprotic (two H+ per molecule)
so it's like 150.0 mL of 0.10 M H+
it takes twice as much NaOH (300.0 mL) to neutralize
so the concentration is half ... 0.05 M ... same as the original acid
To find the concentration of the sodium hydroxide solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH). The balanced equation is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
From the balanced equation, we can see that the mole ratio between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is 1:2.
First, let's calculate the number of moles of sulfuric acid:
Moles of H2SO4 = volume of H2SO4 solution (in liters) x concentration of H2SO4 solution
Volume of H2SO4 solution = 150.0 mL = 0.1500 L
Concentration of H2SO4 solution = 0.05 M
Moles of H2SO4 = 0.1500 L x 0.05 mol/L = 0.0075 mol
Since the mole ratio between H2SO4 and NaOH is 1:2, the number of moles of sodium hydroxide is twice the number of moles of sulfuric acid:
Moles of NaOH = 2 x Moles of H2SO4 = 2 x 0.0075 mol = 0.015 mol
Now, let's calculate the concentration of the sodium hydroxide solution:
Concentration of NaOH = Moles of NaOH / Volume of NaOH solution (in liters)
Volume of NaOH solution = 300.0 mL = 0.3000 L
Concentration of NaOH = 0.015 mol / 0.3000 L = 0.05 M
Therefore, the concentration of the sodium hydroxide solution is 0.05 M, which does not match the answer you provided (0.25 M). Please double-check your calculations or provide additional information if necessary.
To find the concentration of the sodium hydroxide (NaOH) solution, we can use the concept of stoichiometry based on the balanced chemical equation for the neutralization reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH):
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
From the equation, we can see that we need two moles of NaOH to neutralize one mole of H2SO4.
1 mole of H2SO4 is equal to its molar concentration multiplied by its volume in liters:
moles of H2SO4 = concentration of H2SO4 * volume of H2SO4 solution (in liters)
= 0.05 M * 0.150 L
= 0.0075 moles
Since a 1:2 stoichiometric ratio is used for the neutralization reaction, we have:
moles of NaOH = 2 * moles of H2SO4
= 2 * 0.0075 moles
= 0.015 moles
Now, we can find the concentration of the NaOH solution using its moles and volume in liters:
concentration of NaOH = moles of NaOH / volume of NaOH solution (in liters)
= 0.015 moles / 0.300 L
= 0.05 M
Therefore, the concentration of the sodium hydroxide (NaOH) solution is 0.05 M. It seems that the answer you got, 0.25 M, is not correct.