Would it take more 0.1 M HCl or 0.1 M H2SO4 to neutralize 30 ml of NaOH?
To determine whether it would take more 0.1 M HCl or 0.1 M H2SO4 to neutralize 30 ml of NaOH, we need to consider the stoichiometry of the reactions between these acids and NaOH.
The balanced chemical equation for the reaction between HCl and NaOH is:
HCl + NaOH → NaCl + H2O
The balanced chemical equation for the reaction between H2SO4 and NaOH is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
From the equations, we can see that one molecule of HCl reacts with one molecule of NaOH, while one molecule of H2SO4 reacts with two molecules of NaOH.
Since both HCl and H2SO4 are 0.1 M, it means that for every liter of acid solution, there are 0.1 moles of acid. This allows us to calculate the number of moles of acid present in 30 ml (0.03 L) of each solution.
For HCl:
Moles of HCl = molarity × volume = 0.1 mol/L × 0.03 L = 0.003 moles of HCl
For H2SO4:
Moles of H2SO4 = molarity × volume = 0.1 mol/L × 0.03 L = 0.003 moles of H2SO4
Now, comparing the stoichiometry of the reactions, we can see that it takes one mole of HCl to react with one mole of NaOH (1:1 ratio). Similarly, it takes one mole of H2SO4 to react with two moles of NaOH (1:2 ratio).
Since the number of moles of HCl is the same as the number of moles of H2SO4 (0.003 moles), and assuming complete reaction, it means that both solutions will just be enough to neutralize 30 ml of NaOH.
Hence, it would take an equal amount of 0.1 M HCl or 0.1 M H2SO4 to neutralize 30 ml of NaOH.
To determine which acid, 0.1 M HCl or 0.1 M H2SO4, would require more to neutralize 30 ml of NaOH, you need to calculate the equivalent amounts of each acid.
1. Calculate the number of moles of NaOH:
Moles = Molarity x Volume
Moles of NaOH = 0.1 M x 30 ml = 3 mmol (millimoles)
2. Determine the molar ratio of acid to NaOH:
Since NaOH is a monoprotic base, it requires one mole of acid to neutralize it. Therefore, the molar ratio of acid to NaOH is 1:1.
3. Calculate the required volume of acid needed to neutralize the NaOH:
For HCl:
Moles of HCl = 3 mmol
Volume of HCl = Moles / Molarity
= 3 mmol / 0.1 M
= 30 ml
For H2SO4:
Moles of H2SO4 = 3 mmol
Volume of H2SO4 = Moles / Molarity
= 3 mmol / 0.1 M
= 30 ml
Based on the calculations, it would take the same volume (30 ml) of either 0.1 M HCl or 0.1 M H2SO4 to neutralize 30 ml of NaOH.
Figure it out. Assume M NaOH is 0.1M.
NaOH + HCl ==> NaCl + H2O
millimols NaOH = mL x M = 30 x 0.1 = 3
mmols HCl needed =3.
M = mmols/mL or mL = mmols/M = 3/0.1 = 30 mL HCl
For H2SO4.
2NaOH + H2SO4 --> Na2SO4 + 2H2O
mmols NaOH = 3
mmols H2SO4 needed = 3 x (1 mol H2SO4/2 mmols H2SO4 = 3 x 1/2 = 1.5 mmols H2SO4 needed.
mL = mmols/M = 1.5/0.1 = 15 mL.