Question

if your concentrations of two different acids were the same, would you have to add the same or different amounts of titrant?

Answer 1

Task:

if your concentrations of two different acids were the same, would you have to add the same or different amounts of titrant?

Answer:

No, the amount of the titrant added depends on how many $\mathrm{H}^+$ ions there are in the acid or how many OH⁻ ions there are in the base (titrant).

For example, if we titrate $100\mathrm{mL}0.1\mathrm{M}\mathrm{H}_3\mathrm{PO}_4$ and $100\mathrm{mL}0.1\mathrm{M}\mathrm{HCl}$ with $0.1\mathrm{M}\mathrm{NaOH}$

1) For the $\mathbf{H}_3\mathbf{PO}_4$ we have:

The amount of $\mathrm{H}_3\mathrm{PO}_4$ $n = C(\mathrm{H}_3\mathrm{PO}_4) \times V(\mathrm{H}_3\mathrm{PO}_4) = 0.1\mathrm{M} \times 0.1\mathrm{L} = 0.01\mathrm{mol}$

According to the chemical equation

$$\mathrm{H}_3\mathrm{PO}_4 + 3\mathrm{NaOH} = \mathrm{Na}_3\mathrm{PO}_4 + 3\mathrm{H}_2\mathrm{O}$$

The amount of NaOH needed is $\mathrm{n(NaOH)} = 3 \times \mathrm{n(H_3PO_4)} = 3 \times 0.01 = 0.03\mathrm{mol}$

The volume of NaOH added is $\mathrm{V} = \mathrm{n(NaOH)} / \mathrm{C(NaOH)} = 0.03 / 0.1 = 0.3\mathrm{L} = 300\mathrm{mL}$

2) For HCl we have

The amount of HCl $n = C(\mathrm{HCl}) \times V(\mathrm{HCl}) = 0.1\mathrm{M} \times 0.1\mathrm{L} = 0.01\mathrm{mol}$

According to the chemical equation

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

The amount of NaOH needed is $\mathrm{n(NaOH)} = \mathrm{n(HCl)} = 0.01\mathrm{mol}$

The volume of NaOH added is $\mathrm{V} = \mathrm{n(NaOH)} / \mathrm{C(NaOH)} = 0.01 / 0.1 = 0.1\mathrm{L} = 100\mathrm{mL}$

As we can see for the same amount of different acids with equal concentrations we have different amount of base (titrant).