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Titolazione di Acido Forte con Base Forte<br />

Si supponga di dover stimare la variazione di pH in 100 mL di HCl 0.1<br />

M per aggiunta graduale di NaOH 0.1 M. In soluzione, hanno luogo:<br />

HCl (aq)<br />

+ H 2<br />

O (l)<br />

H 3<br />

O + (aq) + Cl- (aq)<br />

NaOH (aq)<br />

Na + (aq) + OH- (aq)<br />

OH - (aq) + H 3 O+ (aq)<br />

2 H 2<br />

O (l)<br />

K = (K -1 14 -2<br />

W ) = 10 M<br />

1) V(NaOH) = 0 mL<br />

[H 3 O + ] = 0.1 M pH = 1<br />

2) V(NaOH) = 10 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 – 0.1×10 mmol = 9 mmol<br />

V TOT = V 0 + V(NaOH) = 110 mL<br />

[H 3 O + ] = n(H 3 O + )/V TOT = 9/110 = 0.0818 M pH = 1.09

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