1.Calculate the mass of sodium hydroxide needed to be weighed to prepare 500 ml of a 0.1 M NaOH solution ?

Question

2. The NaOH solution in question 1, was prepared. It’s then standardized by titrating it against 0.5688g of potassium acid phthalate, which is a primary standard acidic substance (KHP = molar mass 204.23 g/mol), according to reaction below. NaOH + HOOCC6H4COOK → NaOOCC6H4COOK + H2O Calculate the exact concentration of NaOH, if 23.06 ml of it was used for end-point titration?

Answer 1

In a titration where the titrant is added to reach an endpoint, moles of acid = moles of base. So, moles of KHP = moles of NaOH at the endpoint.

If 0.5688 g of KHP is being titrated to an endpoint then you would obtain moles of NaOH.

0.5688 g KHP x 1 mol/ (204.23 g KHP) = 2.785 x 10-3 moles KHP

The acid-base reaction should be evaluated to see if it is 1:1 acid/base:

1 eq NaOH + 1 eq KHP -> NaKP + H2O

KHP actually stands for potassium hydrogen phthlate and is able to lose only 1 proton per molecule. Therefore it is 1 to 1 with NaOH.

2.785 x 10-3 moles KHP is neutralized by 2.785 x 10-3 moles of NaOH since the above equation is 1 to 1.

The question states that 23.06 mL of NaOH solution was used to titrated KHP to the endpoint.

Thus, 2.785 x 10-3 moles NaOH is contained in 23.06 mL of NaOH solution.

The question also asks for the concentration of the NaOH solution used to neutralize the KHP.

M (molarity) is expressed in moles per liter (moles/L).

So, 23.06 mL can be expressed as 0.02306 L.

2.785 x 10 -3 moles NaOH/ 0.02306 L = 0.121 M NaOH