Question

An unknown diprotic acid (H2A) requires 44.391 mL of 0.111 M NaOH to completely neutralize a 0.58 g sample. Calculate the approximate molar mass of the acid.

Answer 1

Answer:

$M=235.42g/mol$

Explanation:

Hello!

In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

$2n_{acid}=n_{base}$

It means that the moles of acid can be computed given the volume and concentration of NaOH:

$n_{acid}=\frac{M_{base}V_{base}}{2} =\frac{0.044391L*0.111mol/L}{2} \\\\n_{acid}=2.46x10^{-4}mol$

It means that the approximate molar mass of the acid is:

$M=\frac{m_{acid}}{n_{acid}} \\\\M=\frac{m_{acid}}{n_{acid}} =\frac{0.58g}{2.46x10^{-3}mol}\\\\M=235.42g/mol$

Best regards!

Answer 2

44.391 mL of 0.111 M NaOH completely neutralize 0.58 g of a diprotic acid, whose molar mass is 2.3 × 10² g/mol.

An unknown diprotic acid is completely neutralized with NaOH.

What is neutralization?

Neutralization is a reaction between an acid and a base to form salt and water.

• Step 1: Write the balanced generic neutralization reaction.

H₂A + 2 NaOH ⇒ Na₂A + 2 H₂O

• Step 2: Calculate the reacting moles of NaOH.

44.391 mL of 0.111 M NaOH react.

(44.391 × 10⁻³ L) × (0.111 mol/L) = 4.93 × 10⁻³ mol

• Step 3: Calculate the reacting moles of H₂A.

The molar ratio of H₂A to NaOH is 1:2.

4.93 × 10⁻³ mol NaOH × (1 mol H₂A/2 mol NaOH) = 2.47 × 10⁻³ mol H₂A

• Step 4: Calculate the molar mass of H₂A

2.47 × 10⁻³ moles of H₂A have a mass of 0.58 g.

M = 0.58 g/(2.47 × 10⁻³ mol) = 2.3 × 10² g/mol

44.391 mL of 0.111 M NaOH completely neutralize 0.58 g of a diprotic acid, whose molar mass is 2.3 × 10² g/mol.

Learn more about neutralization here: https://brainly.com/question/203541