2H2O(l)⇌H3O++HO−
As for any equilibrium, we can write the equilibrium expression:
[H3O+][HO−][H2O] = K'w
Since [H2O] is huge, we can simplify this expression:
[H3O+][HO−] = Kw
At 298K, [H3O+][HO−] = Kw=10−14
Taking log10 of both sides:
log10[H3O+]+log10[HO−] = log10(10−14) = −14
Now we can define pH = −log10[H3O+], and
pOH = −log10[HO−].
And thus pOH+pH=14 (finally!)
Thus pOH of 1×10−4mol⋅L−1 NaOH = −log10(1×10−4)=4
Since pOH+pH=14, pH=10, as required.
[3]pH is calculated using Handerson- Hasselbalch equation,
pH = pKa + log [conjugate base] / [acid]
Conjugate Base = Acetate (CH₃COO⁻)
Acid
= Acetic acid (CH₃COOH)
So,
pH = pKa + log [acetate] / [acetic acid]
We are having conc. of acid and acetate but missing with pKa,
pKa is calculated as,
pKa = -log Ka
Putting value of Ka,
pKa = -log 1.76 × 10⁻⁵
pKa = 4.75
Now,
Putting all values in eq. 1,
pH = 4.75 + log [0.172] / [0.818]
pH = 4.072