pH fun (1 Viewer)

[Julz666]

Could someone please show me how to do this:
20mL of 0.08 mol/L HCl is added to 30mL of 0.05 mol/L of NaOH
Calculate the pH of the solution

Thanks

 

[Gedi-Master]

Hey, im not 100% sure but ill give it a go

HCl + NaOH -> NaCl + H2O

Therefore 1 mol of HCl reacts with 1 mol of NaOH

Now calculate the number of moles of each

n=cV

HCl:

c= 0.08 mol/L

V= 0.02L

Therefore number of moles of HCl = 0.0016 mol

NaOH:

c= 0.05 mol/L

V= 0.03L

Therefore number of moles of NaOH = 0.0015 mol

Thus due to the mol ratio being 1:1,

you will have 0.0015 mol of the
0.0016 mol of the HCl

reacting with

0.0015 mol (all of it) of the NaOH

This means that you will have an excess 0.0001 mol of the HCl remaining in the now nuetralised solution

therefore you must find the concentration of the excess HCl in the solution,

c= n/V

n= 0.0001 mol

V(of the solution)= 0.05L

Therefore the concentration of the excess HCl in the solution= 0.002 mol/L

Now you need to find the pH of the solution...

therefore you need to calculate it using the concentration of the excess HCl in the solution,

0.002 mol/L

-log(0.002)

which will tell you that the pH of the solution is 2.69, or 2.7

 

[Julz666]

Test No. 10008896 : Results - Online Multiple Choice

Sadly they wanted an answer of 2.7

 

[Gedi-Master]

damn it! sorry,

what did u get?

 

[Julz666]

What I did was subtracted the moles of H+ from moles of OH-, giving an answer of 0.0001 moles of H+ left over. I then stupidly went straight to the pH equation and got -log(0.0001) and got an answer of 4 out, which is wrong

 

[Gedi-Master]

YES!!!

i found out where i made my mistake!!!!!

ill edit it

it was when i added the volumes of the solutions, i added the concentrations by mistake!:jaw:

 

[Julz666]

Yup, that works now.
Thanks

 

[Gedi-Master]

Cool, your welcome