Question

1 gram of sodium hydroxide was treated with 25 ml. of 0.75 M HCI solution, the mass of sodium hydroxide left unreacted is equal to

• A. 750mg
• B. 250 mg
• C. Zero mg
• D. 200 mg

Answer 1

The Correct Option is B

Step 1: Write the Reaction 

The neutralization reaction between NaOH and HCl is:

$$ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + H_2O $$

Step 2: Calculate the Moles of NaOH

Given:

• Mass of NaOH = 1 g
• Molar mass of NaOH = 40 g/mol

Moles of NaOH:

$$ \text{Moles of NaOH} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1}{40} = 0.025 \text{ mol} $$

Step 3: Calculate the Moles of HCl

Given:

• Volume of HCl = 25 mL = 0.025 L
• Molarity of HCl = 0.75 M

Moles of HCl:

$$ \text{Moles of HCl} = M \times V = 0.75 \times 0.025 = 0.01875 \text{ mol} $$

Step 4: Determine the Limiting Reagent

The reaction follows a 1:1 molar ratio between NaOH and HCl.

• NaOH available: 0.025 mol
• HCl available: 0.01875 mol

Since HCl has fewer moles, it is the limiting reagent.

Step 5: Calculate Unreacted NaOH

Unreacted moles of NaOH:

$$ \text{Unreacted NaOH} = 0.025 - 0.01875 = 0.00625 \text{ mol} $$

Mass of unreacted NaOH:

$$ \text{Mass} = 0.00625 \times 40 = 0.25 \text{ g} = 250 \text{ mg} $$

Step 6: Conclusion

The mass of unreacted NaOH is 250 mg.