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A 25.0 mL sample of sulfuric acid is titrated with 30.0 mL of 0.150 M sodium hydroxide. Calculate the concentration of the sulfuric acid.

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Answer: The concentration of the sulfuric acid is 0.09 M

Step-by-step explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

According to neutralization law,


n_1M_1V_1=n_2M_2V_2

where,


n_1 = basicity of sulphuric acid
(H_2SO_4) = 2


M_1 = Molarity of sulphuric acid
(H_2SO_4)solution = ?


V_1 = volume of sulphuric acid
(H_2SO_4)solution = 25.0 ml


n_2 = acidity of sodium hydroxide
(NaOH) = 1


M_2 = Molarity of sodium hydroxide
(NaOH) = 0.150 M


V_2 = volume of sodium hydroxide
(NaOH) = 30.0 ml


2* M_1* 25.0=1* 0.150* 30.0


M_1=0.09M

Thus the concentration of the sulfuric acid is 0.09 M

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