pH larutan Penyangga Asam
pH Larutan Penyangga Asam
\([\ce{H+}] = \text{Ka} \:.\: \dfrac {\text{mol asam lemah}}{\text{mol basa konjugasi}}\)
Larutan penyangga asam terdiri atas dua larutan:
- Asam lemah
- Basa konjugasi
Contoh
Larutan yang terdiri atas \(\ce{CH3COOH}\) dan \(\ce{CH3COONa}\)
\begin{equation*}
\begin{array}
& \ce{CH3COOH (aq) & <=> & CH3COO- (aq) & + & H+ (aq)} \quad {\color {red} \dotso \: (1)} \\\\
{\color{blue}\text{asam lemah}} && &&
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}
& \ce{CH3COONa (aq) & -> & CH3COO- (aq) & + & Na+ (aq)} \quad {\color {red} \dotso \: (2)}\\\\
&& {\color{blue}\text{basa konjugasi}} &&
\end{array}
\end{equation*}
Reaksi pertama merupakan reaksi kesetimbangan dengan tetapan kesetimbangan Ka
\begin{equation*}
\begin{split}
\text{Ka} & = \frac {[\ce{CH3COO-}] \:.\: [\ce{H+}]}{[\ce{CH3COOH}]} \\\\
[\ce{CH3COO-}] \:.\: [\ce{H+}] & = \text{Ka} \:.\: [\ce{CH3COOH}] \\\\
[\ce{H+}] & = \frac {\text{Ka} \:.\: [\ce{CH3COOH}]}{[\ce{CH3COO-}]} \\\\
[\ce{H+}] & = \text{Ka} \:.\: \frac {\ce{mol CH3COOH}}{\ce{mol CH3COO-}}
\end{split}
\end{equation*}
Catatan:
Mol \(\ce{mol CH3COOH}\) berasal dari reaksi 1
Mol \(\ce{mol CH3COO-}\) berasal dari reaksi 2
Mol \(\ce{mol CH3COO-}\) yang berasal dari reaksi 1 nilainya kecil dan diabaikan
pH larutan Penyangga Basa
pH Larutan Penyangga Basa
\([\ce{OH-}] = \text{Kb} \:.\: \dfrac {\text{mol basa lemah}}{\text{mol asam konjugasi}}\)
Larutan penyangga basa terdiri atas dua larutan:
- Basa lemah
- Asam konjugasi
Contoh
Larutan yang terdiri atas \(\ce{NH4OH}\) dan \(\ce{NH4Cl}\)
\begin{equation*}
\begin{array}
& \ce{NH4OH (aq) & <=> & NH4+ (aq) & + & OH- (aq)} \quad {\color {red} \dotso \: (1)} \\\\
{\color{blue}\text{basa lemah}} && &&
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}
& \ce{NH4Cl (aq) & -> & NH4+ (aq) & + & Cl- (aq)} \quad {\color {red} \dotso \: (2)}\\\\
&& {\color{blue}\text{asam konjugasi}} &&
\end{array}
\end{equation*}
Reaksi pertama merupakan reaksi kesetimbangan dengan tetapan kesetimbangan Kb
\begin{equation*}
\begin{split}
\text{Kb} & = \frac {[\ce{NH4+}] \:.\: [\ce{OH-}]}{[\ce{NH4OH}]} \\\\
[\ce{NH4+}] \:.\: [\ce{OH-}] & = \text{Kb} \:.\: [\ce{NH4OH}] \\\\
[\ce{OH-}] & = \frac {\text{Kb} \:.\: [\ce{NH4OH}]}{[\ce{NH4+}]} \\\\
[\ce{OH-}] & = \text{Kb} \:.\: \frac {\ce{mol NH4OH}}{\ce{mol NH4+}}
\end{split}
\end{equation*}
Catatan:
Mol \(\ce{mol NH4OH}\) berasal dari reaksi 1
Mol \(\ce{mol NH4+}\) berasal dari reaksi 2
Mol \(\ce{mol NH4+}\) yang berasal dari reaksi 1 nilainya kecil dan diabaikan