sodium hydroxide and sulphuric acid solutions.

Question

in the reaction between sodium hydroxide and sulphuric acid solutions, what volume of 0.5molar sodium hydroxide would exactly neutralize10cm cube of 1.25 molar sulphuric acid?

Answer

Answer

Answer

50 cm³ of 0.5 m sodium hydroxide is required to exactly neutralize 10 cm³ of 1.25 m sulfuric acid.

Answer

the balanced chemical equation is:
[
\mathrm{h_2so_4 + 2naoh ightarrow na_2so_4 + 2h_2o}
]
genui{"chemistry_reactions_solutions_thermodynamics_learning_block":{"type_id":"acid_base_titration"}}
step 1: convert the volume of sulfuric acid to dm³
[
10\ 	ext{cm}^3 = \frac{10}{1000} = 0.010\ 	ext{dm}^3
]
step 2: calculate the moles of sulfuric acid
[
	ext{moles} = 	ext{concentration} 	imes 	ext{volume}
]
[
= 1.25 	imes 0.010 = 0.0125\ 	ext{mol}
]
step 3: use the mole ratio
from the equation:
1 mole of h₂so₄ reacts with 2 moles of naoh.
so,
[
	ext{moles of naoh} = 2 	imes 0.0125 = 0.025\ 	ext{mol}
]
step 4: calculate the volume of 0.5 m naoh required
[
	ext{volume} = \frac{	ext{moles}}{	ext{concentration}}
]
[
= \frac{0.025}{0.5} = 0.05\ 	ext{dm}^3
]
convert to cm³:
[
0.05 	imes 1000 = 50\ 	ext{cm}^3
]
answer:
50 cm³ of 0.5 m sodium hydroxide is required to exactly neutralize 10 cm³ of 1.25 m sulfuric acid.

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sodium hydroxide and sulphuric acid solutions.

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