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3) V(NaOH) = 20 mL<br />

n(H 3 O + ) = n 0 (H 3 O + ) – n(OH - ) = C 0 (H 3 O + )V(HCl) - C 0 (OH - )V(NaOH) =<br />

= 0.1 × 100 mmol – 0.1 × 20 mmol = 8 mmol<br />

V TOT = V 0 + V(NaOH) = 120 mL<br />

[H 3 O + ] = n(H 3 O + )/V TOT = 8/120 = 0.0667 M pH = 1.18<br />

4) V(NaOH) < 100 mL<br />

n(H 3 O + ) = n 0 (H 3 O + ) – n(OH - ) = C 0 (H 3 O + )V(HCl) - C 0 (OH - )V(NaOH) =<br />

= 0.1× 100 mmol – 0.1 × V(NaOH) mmol<br />

V TOT = 100 + V(NaOH) mL<br />

[H 3 O + ] = n(H 3 O + )/V TOT = [10 - 0.1 × V(NaOH)]/[100 + V(NaOH)]

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