Chemistry problem solving (neutralisation)

[A_I_]

On se propose de préparer une solution de pH= 4.4 en utilisant 100 ml d'une solution d'acide benzoïque de concentration 2,5.10^-2 mol.L^-1
pKa C6H5COOH/C6H5COO- = 4.20

a)quel volume V d'une solution d'hydroxyde de sodium NaOH de concentration 0.1 mol.L^-1 faut-il lui ajouter?

b)si à la place de la solution d'hydroxyde de sodium on utilise une solution de benzoate de sodium de même concentration, quel volume V' de cette solution faut-il ajouter?

I hope someone will answer me :)
thanks :)
JOE

 

[Astronuc]

This might be of help - http://chimge.unil.ch/Fr/ph/1ph34.htm

from - http://chimge.unil.ch/Fr/ph/1ph1.htm

It appears that on is to prepare a solution with pH=4.4 using 0.1 L (liter) of 0.025 M benzoic acid (pKa = 4.2),

In part a, one is asked 'what volume, V, of NaOH of 0.1 M.

So moles of NaOH = 0.1 V, and this added to a solution 0.1 L containing 0.0025 moles of C₆H₅COOH.

the resulting volume will be (0.1+V).

Remember, the strong base will react completely with molar equivalent of a weak acid, so not much NaOH is needed.

Perhaps then you can solve b.

 

[wais]

a) To prepare a solution with a pH of 4.4 using 100 ml of a 2.5 x 10^-2 mol/L solution of benzoic acid, we first need to determine the amount of benzoic acid present in the solution. This can be calculated using the formula C1V1 = C2V2, where C1 and V1 are the initial concentration and volume, and C2 and V2 are the final concentration and volume.

Using this formula, we can calculate the amount of benzoic acid present in 100 ml of the solution:

C1V1 = C2V2
(2.5 x 10^-2 mol/L)(100 ml) = (C2)(100 ml)
C2 = 2.5 x 10^-2 mol/L

This means that there is 2.5 x 10^-2 mol of benzoic acid present in 100 ml of the solution.

Next, we need to determine the amount of hydroxide ions (OH-) needed to neutralize this amount of benzoic acid. The neutralization reaction between benzoic acid and sodium hydroxide is:

C6H5COOH + NaOH → C6H5COONa + H2O

From this equation, we can see that 1 mole of benzoic acid reacts with 1 mole of sodium hydroxide to produce 1 mole of water. Therefore, we need 2.5 x 10^-2 mol of sodium hydroxide to neutralize the benzoic acid in the solution.

To calculate the volume of 0.1 mol/L sodium hydroxide needed, we can use the formula C1V1 = C2V2, where C1 = 0.1 mol/L, C2 = 2.5 x 10^-2 mol/L, and V2 = 2.5 x 10^-2 mol.

C1V1 = C2V2
(0.1 mol/L)(V1) = (2.5 x 10^-2 mol/L)(2.5 x 10^-2 mol)
V1 = 0.25 L = 250 ml

Therefore, we need to add 250 ml of 0.1 mol/L sodium hydroxide to the solution of benzoic acid to achieve a pH of 4.4.

b) If