Question

What mass of naoh will react completely with 25.0 ml of 3.00 m h2so4?

Answer 1

$\huge{\boxed{\mathfrak{\red{ANSWER}}}}$

Acid-Base Reaction:

Sodium hydroxide is a strong base and sulphuric acid is a strong acid. Sulphuric acid is a dibasic acid. That means to neutralize one molecule of sulphuric acid two molecules of base are required.

Now,

$molarity = \frac{moles \: of \: h2so4}{volume \: of \: solution}$

Moles of H2SO4 = Volume of solution × Molarity

Moles of H2SO4 = 0.025L × 3mL = 0.075 moles

{1000 ml = 1L}

As sulphuric acid is a dibasic acid,so 2 NaOH molecules are require to neutralise 1 sulphuric acid molecule.

Hence,Moles of NaOH require = 0.075 mol*2 = 0.15 mol.

Mass of NaOH = mole*molar mass = 0.15mol*40g/mol = 6g.

So,6 g of NaOH is required to react exactly with 25.0 ml of 3.0 M H2SO4.

Answer 2

Explanation:

refer above solution

law of chemical equivalence

1eq of h2so4 = 1 eq of naoh