Berikut enam level latihan: eksponen negatif, operasi notasi ilmiah, akar kuadrat, dan pH. Setiap level dilengkapi enam contoh unik beserta langkah perhitungan menggunakan \(\KaTeX\). Artikel ini sangat cocok untuk mengawali pokok bahasan hidrolisis, karena memastikan semua siswa memahami aturan dan pola perhitungan matematika, sehingga pembelajaran hidrolisis garam dapat lebih fokus pada konsep kimia daripada terhambat oleh angka-angka.
L1
Level 1: Makna Eksponen Negatif
Ubah bentuk \(10^{-n}\) dan \(a \times 10^{-n}\) ke desimal, atau sebaliknya. Ini adalah fondasi utama perhitungan hidrolisis.
Contoh 1.1
\(\text{Nyatakan } 10^{-4} \text{ dalam bentuk desimal.}\)
\[
\begin{aligned}
10^{-4} &= \frac{1}{10^{4}} \\
&= \frac{1}{10.000} \\
&= 0{,}0001
\end{aligned}
\]
Contoh 1.2
\(\text{Ubah } 3 \times 10^{-2} \text{ ke desimal.}\)
\[
\begin{aligned}
3 \times 10^{-2} &= 3 \times 0{,}01 \\
&= 0{,}03
\end{aligned}
\]
Contoh 1.3
\(\text{Tulis } 0{,}0007 \text{ dalam notasi ilmiah.}\)
\[
\begin{aligned}
0{,}0007 &= 7 \times 10^{-4}
\end{aligned}
\]
Contoh 1.4
\(\text{Nyatakan } 2{,}5 \times 10^{-3} \text{ sebagai desimal.}\)
\[
\begin{aligned}
2{,}5 \times 10^{-3} &= 0{,}0025
\end{aligned}
\]
Contoh 1.5
\(\text{Tulis } 0{,}00005 \text{ dalam notasi eksponen.}\)
\[
\begin{aligned}
0{,}00005 &= 5 \times 10^{-5}
\end{aligned}
\]
Contoh 1.6
\(\text{Sederhanakan } 10^{-6} + 2 \times 10^{-6}.\)
\[
\begin{aligned}
10^{-6} + 2 \times 10^{-6} &= (1+2) \times 10^{-6} \\
&= 3 \times 10^{-6}
\end{aligned}
\]
L2
Level 2: Perkalian & Pembagian Notasi Ilmiah
Latih pembagian eksponen negatif dan perkalian dengan koefisien. Penting untuk menghitung \(K_h = \dfrac{K_w}{K_a}\) atau \(\dfrac{10^{-14}}{K_b}\).
Contoh 2.1
\(\displaystyle \frac{10^{-5}}{10^{-2}} = ....\)
\[
\begin{aligned}
\frac{10^{-5}}{10^{-2}} &= 10^{-5-(-2)} \\
&= 10^{-5+2} \\
&= 10^{-3}
\end{aligned}
\]
Contoh 2.2
\(\displaystyle \frac{10^{-14}}{10^{-9}} =....\)
\[
\begin{aligned}
\frac{10^{-14}}{10^{-9}} &= 10^{-14-(-9)} \\
&= 10^{-14+9} \\
&= 10^{-5}
\end{aligned}
\]
Contoh 2.3
\(\displaystyle (2 \times 10^{-4}) \times (3 \times 10^{-2}) =....\)
\[
\begin{aligned}
(2 \times 10^{-4}) \times (3 \times 10^{-2})&= (2 \times 3) \times 10^{-4+(-2)} \\
&= 6 \times 10^{-6}
\end{aligned}
\]
Contoh 2.4
\(\displaystyle \frac{1{,}8 \times 10^{-5}}{1{,}2 \times 10^{-2}} =....\)
\[
\begin{aligned}
\frac{1{,}8 \times 10^{-5}}{1{,}2 \times 10^{-2}} &= \frac{1{,}8}{1{,}2} \times 10^{-5-(-2)} \\
&= 1{,}5 \times 10^{-5+2}\\&= 1{,}5 \times 10^{-3}
\end{aligned}
\]
Contoh 2.5
\(\displaystyle \frac{10^{-14}}{2 \times 10^{-5}} =....\)
\[
\begin{aligned}
\frac{1 \times 10^{-14}}{2 \times 10^{-5}}&= \frac{1}{2} \times 10^{-14-(-5)} \\
&= 0{,}5 \times 10^{-14+5} \\
&= 0{,}5 \times 10^{-9} \\
&= 5 \times 10^{-1} \times 10^{-9} \\
&= 5 \times 10^{-1-9}\\
&= 5 \times 10^{-10}
\end{aligned}
\]
Contoh 2.6
\(\displaystyle \left(\frac{10^{-14}}{10^{-6}}\right) \times 0{,}02 =....\)
\[
\begin{aligned}
\left(\frac{10^{-14}}{10^{-6}}\right) \times 0{,}02&= 10^{-8} \times (2 \times 10^{-2}) \\
&= 2 \times 10^{-8+(-2)} \\
&= 2 \times 10^{-10}
\end{aligned}
\]
L3
Level 3: Akar Kuadrat \(10^{-n}\)
Pola: \(\sqrt{10^{-n}} = 10^{-n/2}\). Latihan untuk \(n\) genap dan \(n\) ganjil.
Contoh 3.1
\(\sqrt{10^{-4}} =....\)
\[
\begin{aligned}
\sqrt{10^{-4}} &= 10^{-4/2} \\
&= 10^{-2} \\
&= 0{,}01
\end{aligned}
\]
Contoh 3.2
\(\sqrt{10^{-6}} =....\)
\[
\begin{aligned}
\sqrt{10^{-6}} &= 10^{-6/2} \\
&= 10^{-3} \\
&= 0{,}001
\end{aligned}
\]
Contoh 3.3
\(\sqrt{10^{-3}} =....\) (\(n\) ganjil)
\[
\begin{aligned}
\sqrt{10^{-3}} &= 10^{-3/2} \\
&= 10^{-1{,}5} \\
&= 10^{0{,}5-2} \\
&= 10^{0{,}5} \times 10^{-2}\\
&\approx 3{,}162 \times 10^{-2}
\end{aligned}
\]
Contoh 3.4
\(\sqrt{10^{-5}} =....\) (\(n\) ganjil)
\[
\begin{aligned}
\sqrt{10^{-5}} &= 10^{-5/2} \\
&= 10^{-2{,}5} \\
&= 10^{0{,}5-3}\\
&= 10^{0{,}5} \times 10^{-3} \\
&\approx 3{,}162 \times 10^{-3}
\end{aligned}
\]
Contoh 3.5
\(\sqrt{10^{-8}} =....\)
\[
\begin{aligned}
\sqrt{10^{-8}} &= 10^{-8/2} \\
&= 10^{-4} \\
&= 0{,}0001
\end{aligned}
\]
Contoh 3.6
\(\sqrt{10^{-1}} =....\)
\[
\begin{aligned}
\sqrt{10^{-1}} &= 10^{-0{,}5} \\
&\approx 0{,}3162
\end{aligned}
\]
L4
Level 4: Akar \(\sqrt{a \times 10^{-n}}\)
Teknik: ubah agar eksponen genap, lalu pisahkan \(\sqrt{a} \times 10^{-n/2}\). Sangat berguna saat eksponen di dalam akar ganjil.
Contoh 4.1
\(\sqrt{4 \times 10^{-4}} =....\)
\[
\begin{aligned}
\sqrt{4 \times 10^{-4}} &= \sqrt{4} \times \sqrt{10^{-4}} \\
&= 2 \times 10^{-2} \\
&= 0{,}02
\end{aligned}
\]
Contoh 4.2
\(\sqrt{9 \times 10^{-2}} =....\)
\[
\begin{aligned}
\sqrt{9 \times 10^{-2}} &= \sqrt{9} \times \sqrt{10^{-2}} \\
&= 3 \times 10^{-1} \\
&= 0{,}3
\end{aligned}
\]
Contoh 4.3 – trik genapkan
\(\sqrt{5 \times 10^{-3}} =....\)
\[
\begin{aligned}
\sqrt{5 \times 10^{-3}} &= \sqrt{50 \times 10^{-4}} \\
&= \sqrt{50} \times 10^{-2} \\
&\approx 7{,}071 \times 10^{-2} \\
&= 0{,}07071
\end{aligned}
\]
Contoh 4.4 – trik genapkan
\(\sqrt{2 \times 10^{-5}} =....\)
\[
\begin{aligned}
\sqrt{2 \times 10^{-5}} &= \sqrt{20 \times 10^{-6}} \\
&= \sqrt{20} \times 10^{-3} \\
&= 2\sqrt{5} \times 10^{-3} \\
&\approx 4{,}472 \times 10^{-3}
\end{aligned}
\]
Contoh 4.5
\(\sqrt{1{,}6 \times 10^{-4}} =....\)
\[
\begin{aligned}
\sqrt{1{,}6 \times 10^{-4}} &= \sqrt{1{,}6} \times 10^{-2} \\
&\approx 1{,}265 \times 10^{-2} \\
&= 0{,}01265
\end{aligned}
\]
Contoh 4.6
\(\sqrt{8 \times 10^{-6}} =....\)
\[
\begin{aligned}
\sqrt{8 \times 10^{-6}} &= \sqrt{8} \times 10^{-3} \\
&= 2\sqrt{2} \times 10^{-3} \\
&\approx 2{,}828 \times 10^{-3}
\end{aligned}
\]
L5
Level 5: Menghitung pH dari \([\text{H}^+]\)
Rumus: \(\text{pH} = -\log[\text{H}^+]\). Gunakan sifat \(\log(a \times 10^{-b}) = \log a - b\).
Contoh 5.1
\([\text{H}^+] = 1 \times 10^{-3}\ \text{M}\)
\[
\begin{aligned}
\text{pH} &= -\log[\text{H}^+]\\
\text{pH} &= -\log(10^{-3}) \\
&= -(-3) \\
&= 3
\end{aligned}
\]
Contoh 5.2
\([\text{H}^+] = 4{,}0 \times 10^{-5}\ \text{M}\)
\[
\begin{aligned}
\text{pH} &= -\log[\text{H}^+]\\
\text{pH} &= -\log(4{,}0 \times 10^{-5}) \\
&= -(\log 4{,}0 + \log 10^{-5}) \\
&= -(0{,}6021 - 5) \\
&= 5 - 0{,}6021 \\
&= 4{,}40
\end{aligned}
\]
Contoh 5.3
\([\text{H}^+] = 2{,}5 \times 10^{-4}\ \text{M}\)
\[
\begin{aligned}
\text{pH} &= -\log[\text{H}^+]\\
\text{pH} &= -\log(2{,}5 \times 10^{-4}) \\
&= -(\log 2{,}5 + \log 10^{-4}) \\
&= -(0{,}3979 - 4) \\
&= 4 - 0{,}3979 \\
&= 3{,}60
\end{aligned}
\]
Contoh 5.4
\([\text{H}^+] = 7{,}2 \times 10^{-6}\ \text{M}\)
\[
\begin{aligned}
\text{pH} &= -\log[\text{H}^+]\\
\text{pH} &= -\log(7{,}2 \times 10^{-6}) \\
&= -(\log 7{,}2 + \log 10^{-6}) \\
&= -(0{,}8573 - 6) \\
&= 6 - 0{,}8573 \\
&= 5{,}14
\end{aligned}
\]
Contoh 5.5
\([\text{H}^+] = 3{,}16 \times 10^{-8}\ \text{M}\)
\[
\begin{aligned}
\text{pH} &= -\log[\text{H}^+]\\
\text{pH} &= -\log(3{,}16 \times 10^{-8}) \\
&= -(\log 3{,}16 + \log 10^{-8}) \\
&= -(0{,}50 - 8) \\
&= 8 - 0{,}50 \\
&= 7{,}50
\end{aligned}
\]
Contoh 5.6
\([\text{H}^+] = 1{,}0 \times 10^{-7}\ \text{M}\)
\[
\begin{aligned}
\text{pH} &= -\log[\text{H}^+]\\
\text{pH} &= -\log(10^{-7}) \\
&= -(-7) \\
&= 7
\end{aligned}
\]
L6
Level 6: Simulasi Hidrolisis Garam (Rumus pH Utuh)
Rumus kation: \([\text{H}^+] = \sqrt{\dfrac{K_w}{K_b} \cdot C}\). Rumus anion: \([\text{OH}^-] = \sqrt{\dfrac{K_w}{K_a} \cdot C}\). Latih urutan langkah secara utuh.
Contoh 6.1 – Garam amonium
\(K_b = 1{,}8\times10^{-5},\ C = 0{,}1\ \text{M}\)
\[
\begin{aligned}
[\text{H}^+] &= \sqrt{\frac{K_w}{K_b} \cdot C} \\[10pt]
&= \sqrt{\frac{10^{-14}}{1{,}8\times10^{-5}} \times 0{,}1} \\[10pt]
&= \sqrt{\frac{10^{-15}}{1{,}8\times10^{-5}}} \\[10pt]
&= \sqrt{5{,}556 \times 10^{-11}} \\[10pt]
&= \sqrt{55{,}56 \times 10^{-12}} \\[10pt]
&\approx 7{,}454 \times 10^{-6}\ \text{M} \\[12pt]
\text{pH} &= -\log[\text{H}^+]\\[6pt]
&=-\log(7{,}454 \times 10^{-6})\\[6pt]
&= 6 - \log 7{,}454 \\[6pt]
&\approx 6 - 0{,}872 \\[6pt]
&= 5{,}13
\end{aligned}
\]
Contoh 6.2 – Natrium asetat
\(K_a = 1{,}8\times10^{-5},\ C = 0{,}2\ \text{M}\)
\[
\begin{aligned}
[\text{OH}^-] &= \sqrt{\frac{K_w}{K_a} \cdot C} \\[10pt]
&= \sqrt{\frac{10^{-14}}{1{,}8\times10^{-5}} \times 0{,}2} \\[10pt]
&= \sqrt{\frac{2 \times 10^{-15}}{1{,}8\times10^{-5}}} \\[10pt]
&= \sqrt{1{,}111 \times 10^{-10}} \\[10pt]
&\approx 1{,}054 \times 10^{-5}\ \text{M} \\[12pt]
\text{pOH} &= -\log[\text{OH}^-]\\[6pt]
&= -\log(1{,}054 \times 10^{-5})\\[6pt]
&= 5 - \log 1{,}054 \\[6pt]
&= 5 - 0,023 \\[6pt]
&\approx 4{,}977 \\[12pt]
\text{pH} &= 14 - \text{pOH}\\[6pt]
&= 14 - 4{,}977 \\[6pt]
&= 9{,}02
\end{aligned}
\]
Contoh 6.3 – Hidrolisis kation
\(K_b = 2\times10^{-5},\ C = 0{,}05\ \text{M}\)
\[
\begin{aligned}
[\text{H}^+] &= \sqrt{\frac{K_w}{K_b} \cdot C} \\[10pt]
&= \sqrt{\frac{10^{-14}}{2\times10^{-5}} \times 5\times10^{-2}} \\[10pt]
&= \sqrt{\frac{5\times10^{-16}}{2\times10^{-5}}} \\[10pt]
&= \sqrt{2{,}5 \times 10^{-11}} \\[10pt]
&= \sqrt{25 \times 10^{-12}} \\[10pt]
&= 5 \times 10^{-6}\ \text{M} \\[12pt]
\text{pH} &= -\log[\text{H}^+]\\[6pt]
&=-\log(5 \times 10^{-6})\\[6pt]
&= 6 - \log 5 \\[6pt]
&= 6 - 0,70 \\[6pt]
&\approx 5{,}30
\end{aligned}
\]
Contoh 6.4 – \(\text{NH}_4\text{Cl}\)
\(K_a(\text{NH}_4^+) = 5{,}6\times10^{-10},\ C = 0{,}02\ \text{M}\)
\[
\begin{aligned}
[\text{H}^+] &= \sqrt{K_a \cdot C} \\[10pt]
&= \sqrt{5{,}6\times10^{-10} \times 0{,}02} \\[10pt]
&= \sqrt{1{,}12\times10^{-11}} \\[10pt]
&= \sqrt{11{,}2\times10^{-12}} \\[10pt]
&\approx 3{,}346 \times 10^{-6}\ \text{M} \\[12pt]
\text{pH} &= -\log[\text{H}^+]\\[6pt]
&=-\log(3{,}346 \times 10^{-6})\\[6pt]
&= 6 - \log 3{,}346 \\[6pt]
&= 6 - 0,52 \\[6pt]
&\approx 5{,}48
\end{aligned}
\]
Contoh 6.5 – Eksponen rumit
\(K_a = 4\times10^{-7},\ C = 0{,}03\ \text{M}\)
\[
\begin{aligned}
[\text{OH}^-] &= \sqrt{\frac{K_w}{K_a} \cdot C} \\[10pt]
&= \sqrt{\frac{10^{-14}}{4\times10^{-7}} \times 3\times10^{-2}} \\[10pt]
&= \sqrt{\frac{3\times10^{-16}}{4\times10^{-7}}} \\[10pt]
&= \sqrt{7{,}5 \times 10^{-10}} \\[10pt]
&= \sqrt{7{,}5} \times 10^{-5} \\[10pt]
&\approx 2{,}739 \times 10^{-5}\ \text{M} \\[12pt]
\text{pOH} &= -\log[\text{OH}^-]\\[6pt]
&= -\log(2{,}739 \times 10^{-5})\\[6pt]
&= 5 - \log 2{,}739 \\[6pt]
&= 5 - 0{,}44 \\[6pt]
&\approx 4{,}56 \\[12pt]
\text{pH} &= 14 - \text{pOH}\\[6pt]
&= 14 - 4{,}56 \\[6pt]
&= 9{,}44
\end{aligned}
\]
Contoh 6.6 – \(K_a = 10^{-5}\)
\(K_a = 1{,}0\times10^{-5},\ C = 0{,}01\ \text{M}\)
\[
\begin{aligned}
[\text{OH}^-] &= \sqrt{\frac{K_w}{K_a} \cdot C} \\[10pt]
&= \sqrt{\frac{10^{-14}}{10^{-5}} \times 10^{-2}} \\[10pt]
&= \sqrt{10^{-9} \times 10^{-2}} \\[10pt]
&= \sqrt{10^{-11}} \\[10pt]
&= \sqrt{10 \times 10^{-12}} \\[10pt]
&\approx 3{,}162 \times 10^{-6}\ \text{M} \\[12pt]
\text{pOH} &= -\log[\text{OH}^-]\\[6pt]
&= -\log(3{,}162 \times 10^{-6})\\[6pt]
&= 6 - \log 3{,}162 \\[6pt]
&= 6 - 0{,}50 \\[6pt]
&\approx 5{,}50 \\[12pt]
\text{pH} &= 14 - \text{pOH}\\[6pt]
&= 14 - 5{,}50 \\[6pt]
&= 8{,}50
\end{aligned}
\]
Tips akhir: Selalu sederhanakan eksponen di dalam akar terlebih dahulu (bagi atau genapkan), baru lakukan operasi akar. Gunakan trik menggenapkan eksponen agar hasilnya lebih rapi dan mudah dihitung tanpa kalkulator.
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