Question

A tanker truck carrying 2.01×10³ kg of concentrated sulfuric acid solution tips over and spills its load. The sulfuric acid solution is 95.0% H₂SO₄ by mass and has a density of 1.84 g/mL. Sodium carbonate (Na₂CO₃) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to neutralize 2.01×10³ kg of sulfuric acid solution? Express your answer with the appropriate units.

A 35.00 mL sample of an unknown H₃PO₄solution is titrated with a 0.100 M NaOH solution. The equivalence point is reached when 26.28 mLof NaOH solution is added. What is the concentration of the unknown H₃PO₄ solution? The neutralization reaction is H₃PO₄(aq)+3NaOH(aq)→3H₂O(l)+Na₃PO₄(aq)

Answer 1

95.0 % by mass H₂SO₄ solution means 95.0 g H₂SO₄ is present in 100 g of solution.

i e 1 kg of concentrated H₂SO₄ solution contain 950 g H₂SO₄. We get conversion factor ( 950 g H₂SO₄ / 1 kg H₂SO₄ )

The mass of H₂SO₄ present in 2.01 10 ³ kg concentrated H₂SO₄ solution = 2.01 10 ³ kg H₂SO₄ ( 950 g H₂SO₄ / 1 kg H₂SO₄ )

= 1909500 g H₂SO₄

Now, consider a neutralization reaction H₂SO₄ + Na₂CO₃ Na₂SO₄ + H₂O + CO ₂

From reaction, 1 mol H₂SO₄ 1 mol Na₂CO₃

i e 98.08 g H₂SO₄ 105.99 g Na₂CO₃

1909500 g H₂SO₄ ( 105.99 1909500 / 98.08 ) g Na₂CO₃

1909500 g H₂SO₄ 2063498 g Na₂CO₃

ANSWER : 2.06 10 ³   kg of  Na₂CO₃ are required to neutralize 2.01 10 ³   kg of 95.0 % by mass of concentrated sulfuric acid solution.

PART 2

Consider reaction, H₃PO₄(aq) + 3 NaOH(aq) 3H₂O (l) + Na₃PO₄(aq)

From reaction, stoichiometric ratio = No. of moles of acid / no. of moles of base = 1/3

We have correlation, M _acid V _acid = M _base V _base stoichiometric ratio

M _acid 35.00 ml = 0.100 M 26.28 ml ( 1/3)

M _acid = 0.100 M 26.28 ml ( 1/3) / ( 35.00 ml )

M _acid = 0.0250 M

ANSWER : Concentration of unknown H₃PO₄ solution is 0.0250 M