# Help needed- How do I do this pH neutralisation ?!Watch

#1
Very stuck here.

I have a question which I have no clue what to do .

A mass, m, of sp;od KOH is added to 755cm3 of 0.0120 M HCl. The pH after this addition is 11.60 at 25 C. The volume of the resulting solution is still 755cm3.

1- calculate the number of moles of OH- ions needed to neutralise exactly the H+ ions present in the 755cm3 of 0.0120 HCl.

Firstly I worked out the pH of the HCl which I got to be 1.92.

Then do I do : 7- 1.92? To get 5.08

Then do the kw = [H+][OH-]
and do 5.08 = 1 x 10 -14 (superscript) / OH
OH = 1 x 10-14 / 5.08

0
11 years ago
#2
Why not just calculate the molarity of the KOH solution after the addition. The pH after this addition is 11.60 and as pH + pOH = 14
Then pOH = 2.4

Therefore the OH- ion concentration = 10^-2.4
This is also the concentration of the KOH

Now you know the volume of solution = 755cm3 and the molarity of the KOH so you can work out the number of moles.

Now work out the no of moles of HCl at the beginning = 0.755 x 0.012

As KOH + HCl --> products (i.e. a 1:1 reaction) then KOH needed to neutralise this number of moles of HCl is the same

Therefore total no of moles of KOH added = (0.755 x 10^-2.4) + (0.755 x 0.012)

And from the moles you can work out the mass...
0
X

new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of East Anglia
All Departments Open 13:00-17:00. Find out more about our diverse range of subject areas and career progression in the Arts & Humanities, Social Sciences, Medicine & Health Sciences, and the Sciences. Postgraduate
Wed, 30 Jan '19
• Aston University
Wed, 30 Jan '19
• Solent University
Sat, 2 Feb '19

### Poll

Join the discussion

Remain (1366)
79.7%
Leave (348)
20.3%