A 40.0 mL sample of 0.18 M HCI is titrated with 0.36 M CoHsNH2. Dctermine the pH at these points: At the beginning (before base
is added) a. show work Ph+Lg hox 0-ip 0-30 34 X10-10 K =3. 9x 10 O
1 answer:
Answer:
at the beginning:
pH = 0.745
Explanation:
HCl is a strong acid, so:
0.18 M 0.18 0.18.....equilibrium
before base is added:
∴ [ H3O+ ] ≅ <em>C </em>HCl = 0.18 M
⇒ pH = - Log [ H3O+ ] = - Log ( 0.18 )
⇒ pH = 0.745
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