Question

How many grams of NaOH would be required to neutralize all the acid in 75.0 mL of 0.0900 N H2SO4​

Answer 1

Answer:

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Explanation:

How many grams of NaOH would be required to neutralize all the acid in 75.0 mL of 0.0900 N H2SO4?

(a) 0.540 g

(b) 0.270 g

(c) 1.32 g

(d) 0.660 g

(e) 0.859 g

ANS*:-

(b) 0.270 g

Answer 2

Answer:

0.270 grams of NaOH neutralizes all the acids in 75.0 mL of 0.0900 N H2SO4.

Explanation:

Given,

Normality of H₂SO₄, N = 0.0900 N

Volume of H₂SO₄, V = 75 ml = 0.075 L

number of equivalents, n = 2

The equation for the neutralization reaction between an acid and a base:

acid + base → salt + water

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

moles of NaOH = moles of H₂SO₄

= M × V = N × V/n

= 0.0900 mol/L × 0.0750 L /2

= 0.003375 mol

moles of NaOH = 2 × 0.003375 mol = 0.00675 mol

mass of NaOH = moles of NaOH × molar mass of NaOH = 0.00675 mol × 40.00 g/mol = 0.270 g

Therefore, 0.270 grams of NaOH neutralizes all the acids in 75.0 mL of 0.0900 N H₂SO₄.

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