Larutan penyangga (buffer) dapat mempertahankan nilai pH larutan agar tidak mengalami perubahan yang signifikan akibat penambahan sedikit asam, basa, atau air.
$$\bbox[yellow, 5px, border: 2px solid red] {[\ce{H}^{+}] = \text{K}_a \cdot \dfrac{\text{ mol asam}}{\text{ mol basa konjugasi}}}$$
Sebanyak 50 ml larutan natrium hidroksida 0,4 M dicampur dengan 75 ml larutan asam asetat 0,6 M. Hitunglah pH sebelum dan sesudah kedua larutan bercampur!
Menghitung pH sebelum bercampur
Larutan \(\ce{NaOH}\) 0,4 M
\([\ce{OH}^{-}] = b \cdot \text{M}\)
\([\ce{OH}^{-}] = 1 \cdot (4 \times 10^{-1})\)
\(\text{pOH} = 1\:-\:\log 4\)
\(\text{pH} = 14\:-\:(1\:-\:\log 4) = 13 + \log 4\)
Larutan \(\ce{CH3COOH}\) 0,6 M
\([\ce{H}^{+}] = \sqrt{\text{Ka} \cdot \text{M}}\)
\([\ce{H}^{+}] = \sqrt{1 \cdot 10^{-5}\cdot 6 \cdot 10^{-1}}\)
\([\ce{H}^{+}] = \sqrt{6 \times 10^{-6}}\)
\([\ce{H}^{+}] = \sqrt{6} \times 10^{-3}\)
\(\text{pH} = 3\:-\:\log \sqrt{6}\)
\(\text{pH} = 3\:-\:\dfrac{1}{2}\cdot \log 6\)
Menghitung pH setelah bercampur
|
\(\ce{NaOH}_{(aq)}\) |
+ |
\(\ce{CH3COOH}_{(aq)}\) |
\(\rightarrow\) |
\(\ce{CH3COONa}_{(aq)}\) |
+ |
\(\ce{H2O}_{(l)}\) |
| Mula-mula |
20 mmol |
|
45 mmol |
|
− |
|
− |
| Reaksi |
20 mmol |
|
20 mmol |
|
20 mmol |
|
|
| Sisa |
0 |
|
25 mmol |
|
20 mmol |
|
|
Komponen penyangganya terdiri atas asam lemah \(\ce{CH3COOH}\) dan basa konjugasi \(\ce{CH3COO}^{-}\)
\(\ce{CH3COONa} \rightarrow \ce{CH3COO}^{-} + \ce{Na}^{+}\)
Jumlah mol \(\ce{CH3COO}^{-}\) = jumlah mol \(\ce{CH3COONa}\) = 20 mmol
\([\ce{H}^{+}] = \text{K}_a \cdot \dfrac{\text{ mol asam}}{\text{ mol basa konjugasi}}\)
\([\ce{H}^{+}] = 1 \cdot 10^{-5} \cdot \dfrac{25 \text{ mmol}}{20 \text{ mmol}}\)
\([\ce{H}^{+}] = 1,25 \times 10^{-5}\)
\(\text{pH} = 5\:-\:\log 1,25\)
$$\bbox[yellow, 5px, border: 2px solid red] {[\ce{OH}^{-}] = \text{K}_b \cdot \dfrac{\text{ mol basa}}{\text{ mol asam konjugasi}}}$$
Sebanyak 2,24 liter gas ammonia pada kondisi STP dialirkan ke dalam larutan asam klorida 50 mL dengan konsentrasi 0,01 M. Hitung pH campuran yang terjadi. (Kb ammonia = \(1 \times 10^{-5}\)).
Jumlah mol \(\ce{NH3}\) mula-mula = \(\dfrac{2,24}{22,4} = 0,1 \text{ mol} = 100 \text{ mmol}\)
|
\(\ce{NH3}_{(g)}\) |
+ |
\(\ce{HCl}_{(aq)}\) |
\(\rightarrow\) |
\(\ce{NH4Cl}_{(aq)}\) |
| Mula-mula |
100 mmol |
|
0,5 mmol |
|
− |
| Reaksi |
0,5 mmol |
|
0,5 mmol |
|
0,5 mmol |
| Sisa |
99,5 mmol |
|
0 |
|
0,5 mmol |
Karena yang bersisa \(\ce{NH3}\) maka akan membentuk larutan penyangga basa.
Komponen penyangganya terdiri dari basa lemah \(\ce{NH3}\) dan asam konjugasi \(\ce{NH4}^{+}\)
\(\ce{NH4Cl} \rightarrow \ce{NH4}^{+} + \ce{Cl}^{-}\)
Jumlah mol \(\ce{NH4}^{+}\) = jumlah mol \(\ce{NH4Cl}\) = 0,5 mmol
\([\ce{OH}^{-}] = \text{K}_b \cdot \dfrac{\text{ mol basa}}{\text{ mol asam konjugasi}}\)
\([\ce{OH}^{-}] = 1 \times 10^{-5} \cdot \dfrac{99,5}{0,5}\)
\([\ce{OH}^{-}] = 199 \times 10^{-5}\)
\([\ce{OH}^{-}] = 1,99 \times 10^{-3}\)
\(\text{pOH} = 3\:-\:\log 1,99\)
\(\text{pH} = 14\:-\: (3\:-\:\log 1,99)\)
\(\text{pH} = 11 + \log 1,99\)
