A mixture `NaOH+Na_(2)CO_(3)` required 25mL of 0.1 M HCl using phenolpththalein as the indicator. However, the same amount of the mixture required 30mL of 0.1M HCl when methyl orange was used as the indicator. The molar ration of `NaOH and Na_(2)CO_(3)` in the mixture was:

Options

A. `2:1`
B. `1:2`
C. `4:1`
D. `1:4`

Answer

To solve the problem of finding the molar ratio of NaOH to Na2CO3 in the mixture, we can follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = number of moles of NaOH
- \( y \) = number of moles of Na2CO3

### Step 2: Analyze the Reaction with Phenolphthalein
When phenolphthalein is used as an indicator:
- NaOH is completely neutralized by HCl.
- Na2CO3 is half neutralized, forming NaHCO3 (sodium bicarbonate).

The neutralization reactions are:
1. \( \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)
2. \( \text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow \text{NaHCO}_3 + \text{NaCl} \)

From the problem, we know that 25 mL of 0.1 M HCl is required. We can calculate the millimoles of HCl used:
\[ \text{Millimoles of HCl} = 0.1 \, \text{M} \times 25 \, \text{mL} = 2.5 \, \text{mmol} \]

The total millimoles of base (NaOH and Na2CO3) neutralized is:
\[ x + \frac{y}{2} = 2.5 \quad \text{(Equation 1)} \]

### Step 3: Analyze the Reaction with Methyl Orange
When methyl orange is used:
- Both NaOH and Na2CO3 are completely neutralized.

The neutralization reactions are:
1. \( \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)
2. \( \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{CO}_3 \)

From the problem, we know that 30 mL of 0.1 M HCl is required. We can calculate the millimoles of HCl used:
\[ \text{Millimoles of HCl} = 0.1 \, \text{M} \times 30 \, \text{mL} = 3 \, \text{mmol} \]

The total millimoles of base neutralized is:
\[ x + 2y = 3 \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
We now have two equations:
1. \( x + \frac{y}{2} = 2.5 \)
2. \( x + 2y = 3 \)

From Equation 1, we can express \( x \):
\[ x = 2.5 - \frac{y}{2} \]

Substituting this expression for \( x \) into Equation 2:
\[ (2.5 - \frac{y}{2}) + 2y = 3 \]
\[ 2.5 - \frac{y}{2} + 2y = 3 \]
\[ 2.5 + \frac{3y}{2} = 3 \]
\[ \frac{3y}{2} = 3 - 2.5 \]
\[ \frac{3y}{2} = 0.5 \]
\[ 3y = 1 \quad \Rightarrow \quad y = \frac{1}{3} \approx 0.33 \]

Now substitute \( y \) back into the expression for \( x \):
\[ x = 2.5 - \frac{0.33}{2} \approx 2.5 - 0.165 \approx 2.335 \]

### Step 5: Calculate the Molar Ratio
Now we have:
- \( x \approx 2.335 \)
- \( y \approx 0.33 \)

The molar ratio of NaOH to Na2CO3 is:
\[ \text{Ratio} = \frac{x}{y} = \frac{2.335}{0.33} \approx 7.07 \approx 7:1 \]

### Final Answer
The molar ratio of NaOH to Na2CO3 in the mixture is approximately **7:1**.