10.0 mL of 0.15M of sulfuric acid is titrated with 0.05M of sodium hydroxide. how many mL of sodium hydroxide solution are required to reach the equivalence point?
answered below
To determine the volume of sodium hydroxide solution required to reach the equivalence point, we need to use the balanced chemical equation and apply the concept of stoichiometry.
The balanced equation for the reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
From the balanced equation, we can see that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide. Therefore, the stoichiometric ratio is 1:2.
1 mole of sulfuric acid is present in 0.01 L (10.0 mL) of 0.15 M sulfuric acid solution:
moles H2SO4 = volume (L) x concentration (M)
moles H2SO4 = 0.01 L x 0.15 M
moles H2SO4 = 0.0015 moles
According to the stoichiometry of the balanced equation, 2 moles of sodium hydroxide are required to neutralize 1 mole of sulfuric acid. Therefore, the moles of sodium hydroxide required will be:
moles NaOH = 2 x moles H2SO4
moles NaOH = 2 x 0.0015 moles
moles NaOH = 0.003 moles
Now, let's find the volume of 0.05 M sodium hydroxide solution needed to obtain 0.003 moles of sodium hydroxide:
volume NaOH (L) = moles NaOH / concentration NaOH
volume NaOH (L) = 0.003 moles / 0.05 M
volume NaOH (L) ≈ 0.06 L
To convert the volume to milliliters, we multiply the value by 1000:
volume NaOH (mL) = 0.06 L x 1000 mL/L
volume NaOH (mL) = 60 mL
Therefore, approximately 60 mL of the sodium hydroxide solution is required to reach the equivalence point in the titration.