A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium carbonate). For titration to the phenolphthalein endpoint, 0.200 g of this sample requires 23.98 mL of 0.100 M HCl. An additional 0.700 mL of 0.100 M HCl is required to reach the methyl orange endpoint. What is the percentage of Na2CO3 by mass in the sample?

3.71%

Explanation:

The phenolphthalein endpoint refers to the reactions:

OH⁻ + H⁺ → H₂O

CO₃⁻² + H⁺ → HCO₃⁻

While the methyl orange endpoint to:

HCO₃⁻ + H⁺ → H₂CO₃

So the additional volume required for the second endpoint tells us the amount of HCO₃⁻ species, which in turn is the total amount of Na₂CO₃ in the sample:

0.700 mL * 0.100 M * = 0.07 mmol HCO₃⁻

Now we calculate the mass of Na₂CO₃, using its molecular weight:

0.07 mmol HCO₃⁻ = 0.07 mmol Na₂CO₃

0.07 mmol Na₂CO₃ * 106 mg/mmol = 7.42 mg Na₂CO₃

No calculations using the volume of the first equivalence point are required because the problem already tells us the mass of the sample is 0.200 g.

0.200 g ⇒ 0.200 * 1000 = 200 mg

%Na₂CO₃ = 7.42 mg/200 mg * 100 = 3.71%

the answer is a

photosynthesis