100 mL of HCI + 35 mL of NaOH, colour of methyl orange in the solution will be

Options

A. red
B. yellow
C. can't be predicted
D. methyl orange is not suitable indicator

Answer

To determine the color of methyl orange in a solution of 100 mL of HCl mixed with 35 mL of NaOH, we can follow these steps:

### Step 1: Identify the nature of the reactants
- Hydrochloric acid (HCl) is a strong acid that dissociates completely in water to give H⁺ and Cl⁻ ions.
- Sodium hydroxide (NaOH) is a strong base that also dissociates completely to give Na⁺ and OH⁻ ions.

### Step 2: Calculate the moles of HCl and NaOH
- Assume the concentration of HCl is 1 M (molarity) for simplicity. Therefore, the moles of HCl in 100 mL (0.1 L) is:
\[ \text{Moles of HCl} = 1 \, \text{M} \times 0.1 \, \text{L} = 0.1 \, \text{moles} \]
- Similarly, for NaOH, assuming it is also 1 M, the moles of NaOH in 35 mL (0.035 L) is:
\[ \text{Moles of NaOH} = 1 \, \text{M} \times 0.035 \, \text{L} = 0.035 \, \text{moles} \]

### Step 3: Determine the neutralization reaction
- The neutralization reaction between HCl and NaOH can be represented as:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- From the stoichiometry of the reaction, 1 mole of HCl reacts with 1 mole of NaOH. Therefore, 0.035 moles of NaOH will neutralize 0.035 moles of HCl.

### Step 4: Calculate the remaining moles of HCl
- After the reaction, the remaining moles of HCl will be:
\[ \text{Remaining moles of HCl} = 0.1 \, \text{moles} - 0.035 \, \text{moles} = 0.065 \, \text{moles} \]

### Step 5: Determine the acidity of the solution
- Since there are still 0.065 moles of HCl remaining, the solution will be acidic.

### Step 6: Analyze the effect of methyl orange
- Methyl orange is an acid-base indicator that turns red in acidic solutions and yellow in basic solutions. Since the solution is acidic due to the remaining HCl, the color of methyl orange will be red.

### Conclusion
The color of methyl orange in the solution will be **red**.

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