# Homework Help: Chemistry problem solving (neutralisation)

1. Jan 2, 2005

### 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

2. Jan 4, 2005

### Astronuc

Staff Emeritus
This might be of help - http://chimge.unil.ch/Fr/ph/1ph34.htm [Broken]

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

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.

Last edited by a moderator: May 1, 2017