i do not understand how to solve this problem, please help
10.0mL 0.5M sulfuric acid can neutralize what volume of 0.5M ammonium hydroxide?
H2SO4 + 2NH4OH -> (NH4)2SO4 + 2H2O
so, 1 mole of H2SO4 reacts with 2moles NH4OH
So, how many moles in 10ml of .5M H2SO4?
.01L * .5moles/L = .05 moles H2SO4
So, how many moles of NH4OH will be used?
I might point out here that it has been shown that NH4OH does not exist. It is convenient to write NH3(aq) + H2O ==> NH4OH ==> NH4^+ + OH^- but the intermediate is not there.
To solve this problem, you need to use the concept of molarity and the balanced chemical equation.
First, let's write the balanced chemical equation for the neutralization reaction between sulfuric acid (H2SO4) and ammonium hydroxide (NH4OH):
H2SO4 + 2NH4OH -> (NH4)2SO4 + 2H2O
From the balanced equation, you can see that 1 mol of sulfuric acid reacts with 2 moles of ammonium hydroxide. This means that the mole ratio between sulfuric acid and ammonium hydroxide is 1:2.
Now, let's calculate the moles of sulfuric acid you have:
Moles of sulfuric acid (H2SO4) = Molarity × Volume (in liters)
= 0.5 M × 10.0 mL / 1000 mL
= 0.005 moles
Since the mole ratio between sulfuric acid and ammonium hydroxide is 1:2, you can calculate the moles of ammonium hydroxide that will be neutralized:
Moles of ammonium hydroxide = 2 × Moles of sulfuric acid
= 2 × 0.005 moles
= 0.01 moles
Finally, you can calculate the volume of 0.5 M ammonium hydroxide that can neutralize these moles:
Volume of ammonium hydroxide = Moles of ammonium hydroxide / Molarity
= 0.01 moles / 0.5 M
= 0.02 L (or 20.0 mL)
Therefore, 10.0 mL of 0.5 M sulfuric acid can neutralize 20.0 mL of 0.5 M ammonium hydroxide.