Question

Why does the titration of 50 mL of 0.1 M H2SO4 require a larger titrant than the titration of 50 mL of 0.1 M HCl? Assume that the titrant being used is 0.1 M NaoH.

Answer 1

Question #47871 – Chemistry – Other

Question:

Why does the titration of 50 mL of 0.1 M H₂SO₄ require a larger titrant than the titration of 50 mL of 0.1 M HCl? Assume that the titrant being used is 0.1 M NaOH.

Answer:

The titration of HCl is described with the equation:

$$\mathrm{NaOH} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{H_2O}$$

The amount of moles of 0.1 M HCl in 50 ml is:

$$\nu(\mathrm{HCl}) = C(\mathrm{HCl}) \times V(\mathrm{HCl}) = 0.1\,\mathrm{M} \times \frac{50\,\mathrm{ml}}{1000} = 5 \cdot 10^{-3}\,\mathrm{moles}$$

The amount of moles of NaOH required to neutralize 5·10⁻³ moles of HCl according to the equation above is the same:

$$\nu_1(\mathrm{NaOH}) = \nu(\mathrm{HCl}) = 5 \cdot 10^{-3}\,\mathrm{moles}$$

The volume of 0.1 M NaOH required for the titration is

$$V_1(\mathrm{NaOH}) = \frac{\nu_1(\mathrm{NaOH})}{C(\mathrm{NaOH})} = \frac{5 \cdot 10^{-3}\,\mathrm{moles}}{0.1\,\mathrm{M}} = 5 \cdot 10^{-2}\,\mathrm{L} = 50\,\mathrm{ml}$$

The equation reaction for H₂SO₄ is:

$$2\,\mathrm{NaOH} + \mathrm{H_2SO_4} \rightarrow \mathrm{Na_2SO_4} + 2\,\mathrm{H_2O}$$

The amount of moles of 0.1 H₂SO₄ in 50 ml of its solution is

$$\nu(\mathrm{H_2SO_4}) = C(\mathrm{H_2SO_4}) \times V(\mathrm{H_2SO_4}) = 0.1\,\mathrm{M} \times \frac{50\,\mathrm{ml}}{1000} = 5 \cdot 10^{-3}\,\mathrm{moles}$$

According to the above equation, two moles of NaOH are required to neutralize one mole of NaOH. Then the amount of NaOH that is required to neutralize 5·10⁻³ moles is

$$\nu_2(\mathrm{NaOH}) = 2 \times \nu(\mathrm{H_2SO_4}) = 2 \times 5 \cdot 10^{-3}\,\mathrm{moles} = 10 \cdot 10^{-3}\,\mathrm{moles}$$

The volume of 0.1 M NaOH that contains 10·10⁻³ moles is

$$V_2(\mathrm{NaOH}) = \frac{\nu_2(\mathrm{NaOH})}{C(\mathrm{NaOH})} = \frac{10 \cdot 10^{-3}\,\mathrm{moles}}{0.1\,\mathrm{M}} = 10 \cdot 10^{-2}\,\mathrm{L} = 100\,\mathrm{ml}$$

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