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?
Look up the concentration term Normality.
Nb*volumebase=Na*Volumeacid
.05*volumebase=.30*10ml
you have posted the same question several times, and it has been amswered before.
To determine how many mL of sodium hydroxide solution are required to reach the equivalence point, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction.
The balanced chemical equation for the neutralization reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is:
H2SO4 + 2NaOH ⟶ Na2SO4 + 2H2O
From the balanced equation, we can see that the stoichiometric ratio between sulfuric acid and sodium hydroxide is 1:2. This means that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide.
Now, let's calculate the number of moles of sulfuric acid given the concentration (Molarity) and volume:
Moles of sulfuric acid = Molarity × Volume (in liters)
= 0.15 M × 0.01 L
= 0.0015 moles
Since the stoichiometric ratio between sulfuric acid and sodium hydroxide is 1:2, we need twice as many moles of sodium hydroxide to reach the equivalence point. Therefore, the number of moles of sodium hydroxide needed is:
Moles of sodium hydroxide = 2 × Moles of sulfuric acid
= 2 × 0.0015 moles
= 0.003 moles
Now, let's calculate the volume of sodium hydroxide solution needed given its concentration:
Volume (in liters) = Moles / Molarity
= 0.003 moles / 0.05 M
= 0.06 L
Finally, convert the volume from liters to milliliters:
Volume (in mL) = Volume (in liters) × 1000
= 0.06 L × 1000
= 60 mL
Therefore, 60 mL of sodium hydroxide solution are required to reach the equivalence point in this titration.