Solution check for pH of Strong Acid and Strong Base

[pavadrin]

Homework Statement

Calculate the pH after 24 mL of 1.00 M NaOH is added from the burette to a beaker containing 25.0 mL of 1.00 M HCl

Homework Equations

pH = -log [H+]
1*10^-14 = [H+][OH-]

The Attempt at a Solution

combined volume = 49mL

n.unreacted(H+) = 0.025-0.024 = 0.001 mol

[H+] = n/V = 0.001/0.049 = 0.02 mol L-1

pH = -log[H+] = 1.69 (2 d.p)



is that correct?

many thanks,
pavadrin

 

[symbolipoint]

Yes.

 

[pavadrin]

okay thank you for the reply