Question

Titration reveals that 11.6 mL of 3.0M sulfuric acid are required to neutralize the sodium hydroxide in 25.00mL of NaOH solution. What is the molarity of the NaOH solution?

Answer 1

Answer:

Explanation:

Molarity of acid(volume of acid)(# of H ions)= molarity of base(volume of base)(# of OH ions)

M(v)(#)=M(v)(#)

sulfuric acid    sodium hydroxide

H2SO4           NaOH

(3)(11.6)(2)=M(25)(1)

M=2.784

Answer 2

The molarity of the NaOH solution is 2.784 M

We'll begin by writing the balanced equation for the reaction.

2NaOH + H₂SO₄ —> Na₂SO₄ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, H₂SO₄ ($n_{a}$) = 1

The mole ratio of the base, NaOH ($n_{b}$) = 2

From the question given above, the following data were obtained:

Volume of acid, H₂SO₄ ($V_{a}$) = 11.6 mL

Molarity of acid, H₂SO₄ ($M_{a}$) = 3 M

Volume of base, NaOH ($V_{b}$) = 25 mL

Molarity of base, ($M_{b}$) =?

$\frac{M_{a} V_{a}}{M_{b} V_{b}} = \frac{n_{a}}{n_{a}}\\\\ \frac{3 (11.6)}{M_{b} (25)} = \frac{1}{2}$

Cross multiply

$M_{b}$ × 25 × 1 = 3 × 11.6 × 2

$M_{b}$ × 25 = 69.6

Divide both side by 25

$M_{b} = \frac{69.6}{25}\\ M_{b} = 2.784 M$

Therefore, the molarity of the NaOH solution is 2.784 M

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