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? Express your answer to three significant figures and include the appropriate units.

1.86% Na₂CO₃

Explanation:

The additional volume that is required to reach endpoint with methyl orange indicator is due the presence of carbonate (Calculated as sodium carbonate). The reaction is:

Na₂CO₃ + 2HCl → H₂O + CO₂ + 2NaCl

Thus, we need to obtain moles of HCl and, using the reaction, solve for moles of sodium carbonate, its mass and percentage using molar mass of Na₂CO₃: 106g/mol.

Moles HCl:

0.700mL = 0.700x10⁻³L * (0.100mol / L) = 7.00x10⁻⁵moles HCl

Moles Na₂CO₃:

7.00x10⁻⁵moles HCl * (1mol Na₂CO₃ / 2 mol HCl) = 3.5x10⁻⁵ moles Na₂CO₃

Mass:

3.5x10⁻⁵ moles Na₂CO₃ * (106g / mol) = 0.00371g Na₂CO₃

Percentage:

0.00371g Na₂CO₃ / 0.200g * 100 =

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.