If you titrated 50 mL of HCl and 50 mL H2SO4, each of the same molarity, with NaOH. What is the numerical ratio of the volume of NaOH required for HCl compared to that for H2SO4?
I assume you want to titrate BOTH H ions of the H2SO4. Let's just make the problem easier by assuming 0.1M HCl, 0.1M H2SO4, and 0.1M NaOH.
mmols HCl = 50 x 0.1 = 5
mmols H2SO4 = 50 x 0.1 to titrate the first H and another 5 to titrate the second so 10 mmols for H2SO4.
Now if you titrate the HCl with 0.1M NaOH it will take 50 mL and if you titrate the H2SO4 with 0.1M NaOH, it takes 50 mL for the first H and another 50 for the second so 100 mL for the H2SO4. So the ratio of VHCl.VH2SO4is .....
To find the numerical ratio of the volume of NaOH required for HCl compared to that for H2SO4, we need to use the concept of stoichiometry.
The balanced chemical equation for the neutralization reaction between an acid and a base is:
Acid + Base -> Salt + Water
In this case, the acids are HCl and H2SO4, and the base is NaOH. The balanced equation for the reaction is as follows:
HCl + NaOH -> NaCl + H2O
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
From the balanced equations, we can see that for every 1 mole of HCl, we need 1 mole of NaOH, and for every 1 mole of H2SO4, we need 2 moles of NaOH.
Since the acids are of the same molarity, they contain the same number of moles. Therefore, the ratio of the volume of NaOH required for HCl compared to that for H2SO4 can be determined by comparing the stoichiometric coefficients in the balanced equations.
For HCl, the ratio is 1 mole NaOH : 1 mole HCl, and for H2SO4, the ratio is 2 moles NaOH : 1 mole H2SO4.
However, since the volume of the solutions is given (50 mL each), we can assume that the molarity of both HCl and H2SO4 is the same. Therefore, the number of moles of both acids is equal, and thus the ratio of the volume of NaOH required for HCl compared to that for H2SO4 is also 1 : 2.
In numerical terms, this means that the volume of NaOH required for HCl is half the volume required for H2SO4.