What is the sum of magic numbers of all solutions gives below : (Interger value is between 50 and 60)
(Magic numver of a solution `=pH` of solution `xx` Weight factor)
`{:(,"Solution",,"Weight factor"),("I",underset((K_(a)=10^(-10)))(0.1 "M HCN"),,2),("II",underset(underset((K_(a)=10^(-5)))(+0.1 M CH_(3) COONa))(0.1 M CH_(3) COOH),,1),("III",0.1 M HCl,,3),("IV",0.1 M NH_(4)OH (K_(b)=10^(-5)),,2),("V",0.01 M NaOH,,0.5),("VI",underset(+10 "ml of 0.1 M" NH_(4)OH)(10 "mL of" 0.01 M CH_(3)COOH),,1):}`.

Explanation

To find the sum of the magic numbers of all the given solutions, we will calculate the pH for each solution and then multiply it by the corresponding weight factor. Finally, we will sum all the magic numbers.

### Step-by-Step Solution:

1. **Solution I: 0.1 M HCN (Ka = 10^(-10))**
- Calculate pKa:
\[ pKa = -\log(10^{-10}) = 10 \]
- Calculate pH:
\[ pH = \frac{1}{2} \cdot (pKa - \log C) = \frac{1}{2} \cdot (10 - \log(0.1)) = \frac{1}{2} \cdot (10 - (-1)) = \frac{1}{2} \cdot 11 = 5.5 \]
- Magic number:
\[ \text{Magic Number} = pH \times \text{Weight Factor} = 5.5 \times 2 = 11 \]

2. **Solution II: 0.1 M CH3COOH and 0.1 M CH3COONa (Ka = 10^(-5))**
- Calculate pKa:
\[ pKa = -\log(10^{-5}) = 5 \]
- Since this is a buffer solution, use the Henderson-Hasselbalch equation:
\[ pH = pKa + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) = 5 + \log\left(\frac{0.1}{0.1}\right) = 5 + 0 = 5 \]
- Magic number:
\[ \text{Magic Number} = 5 \times 1 = 5 \]

3. **Solution III: 0.1 M HCl**
- Calculate pH:
\[ pH = -\log(0.1) = 1 \]
- Magic number:
\[ \text{Magic Number} = 1 \times 3 = 3 \]

4. **Solution IV: 0.1 M NH4OH (Kb = 10^(-5))**
- Calculate pKb:
\[ pKb = -\log(10^{-5}) = 5 \]
- Calculate pOH:
\[ pOH = pKb + \log\left(\frac{C}{2}\right) = 5 + \log(0.1/2) = 5 - 0.301 = 4.699 \approx 4.7 \]
- Calculate pH:
\[ pH = 14 - pOH = 14 - 4.7 = 9.3 \]
- Magic number:
\[ \text{Magic Number} = 9.3 \times 2 = 18.6 \approx 19 \]

5. **Solution V: 0.01 M NaOH**
- Calculate pOH:
\[ pOH = -\log(0.01) = 2 \]
- Calculate pH:
\[ pH = 14 - pOH = 14 - 2 = 12 \]
- Magic number:
\[ \text{Magic Number} = 12 \times 0.5 = 6 \]

6. **Solution VI: 10 mL of 0.01 M CH3COOH and 10 mL of 0.1 M NH4OH**
- Since it’s a weak acid and weak base mixture, the pH will be approximately 7.
- Magic number:
\[ \text{Magic Number} = 7 \times 1 = 7 \]

### Final Calculation of the Sum of Magic Numbers:
\[ \text{Total Magic Number} = 11 + 5 + 3 + 19 + 6 + 7 = 51 \]

### Answer:
The sum of the magic numbers of all solutions is **51**.