What volume of 0.7 molar sodium hydroxide is needed to completely neutralize 90.04 mL of 0.4 molar sulfuric acid?
NormalityAcid*Vacid=Normalitybase*Vbase
.8*90.04ml=.7*Vb
solve for Vb
If you don't understand the titration equation,or the concept of Normal solutions, ...
http://dwb.unl.edu/Teacher/NSF/C12/C12Links/users.ev1.net/7Evklawinski/chpt24ntitration.html
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To find the volume of sodium hydroxide needed to neutralize the sulfuric acid, you would use the stoichiometry of the balanced chemical equation between sodium hydroxide (NaOH) and sulfuric acid (H2SO4).
The balanced chemical equation is:
2 NaOH + H2SO4 → Na2SO4 + 2 H2O
From the equation, you can see that one mole of sulfuric acid reacts with two moles of sodium hydroxide.
First, calculate the number of moles of sulfuric acid:
Moles H2SO4 = Volume (L) × Molarity (mol/L)
Moles H2SO4 = 0.09004 L × 0.4 mol/L
Moles H2SO4 = 0.036016 mol
According to the stoichiometry, the number of moles of sodium hydroxide required to neutralize the sulfuric acid would be twice the number of moles of sulfuric acid.
Moles NaOH = 2 × Moles H2SO4
Moles NaOH = 2 × 0.036016 mol
Moles NaOH = 0.072032 mol
Now, to find the volume of sodium hydroxide, use its molarity:
Volume NaOH = Moles NaOH / Molarity (mol/L)
Volume NaOH = 0.072032 mol / 0.7 mol/L
Volume NaOH ≈ 0.1029 L or 102.9 mL
Therefore, approximately 102.9 mL of 0.7 Molar sodium hydroxide is needed to completely neutralize 90.04 mL of 0.4 Molar sulfuric acid.