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?
M1=0.05M
M2=?
V1=150 mL
V2=300mL
So,
M1V1=M2V2
(0.05)(150)=(300)M2
M2(300)=7.5
M2=7.5/300
M2=0.025M
To find the concentration of the sodium hydroxide (NaOH) solution, you need to use the concept of stoichiometry and the balanced equation for the reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH).
The balanced equation for the reaction is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
In this reaction, 1 mole of sulfuric acid (H2SO4) reacts with 2 moles of sodium hydroxide (NaOH).
Given the volume and concentration of the sulfuric acid solution, we can calculate the number of moles of sulfuric acid:
Moles of H2SO4 = Volume (in L) x Concentration (in mol/L)
Moles of H2SO4 = 150.0 mL x 0.05 mol/L
Moles of H2SO4 = 0.0075 mol
Since the stoichiometric ratio between sulfuric acid and sodium hydroxide is 1:2, the number of moles of sodium hydroxide will be twice the moles of sulfuric acid:
Moles of NaOH = 2 x Moles of H2SO4
Moles of NaOH = 2 x 0.0075 mol
Moles of NaOH = 0.015 mol
Now, we can determine the concentration of the sodium hydroxide solution using the volume of the sodium hydroxide solution:
Concentration of NaOH (in mol/L) = Moles of NaOH / Volume (in L)
Concentration of NaOH = 0.015 mol / 0.300 L
Concentration of NaOH = 0.05 mol/L or 0.05 M
Therefore, the concentration of the sodium hydroxide (NaOH) solution is 0.05 M.