Question 8 0.415 g of a pure acid, H₂X, is added to exactly 100 mL of 0.105 M NaOH(aq). A reaction occurs according to the equation H₂X(s) + 2NaOH(aq) → Na₂X(aq) + 2H₂O(l) The NaOH is in excess. This excess NaOH requires 25.21 mL of 0.197 M HCl(aq) for neutralisation. Calculate i. the amount, in mol, of NaOH that is added to the acid H₂X initially. ii. the amount, in mol, of NaOH that reacts with the acid H₂X. iii. the molar mass, in g mol⁻¹, of the acid H₂X.

SSCE VCE Chemistry Acid-Base Reactions: Question 8 0.415 g of a pure ac

Question 8 0.415 g of a pure acid, H₂X, is added to exactly 100 mL of 0.105 M NaOH(aq) - VCE - SSCE Chemistry - Question 8 - 2008 - Paper 1

Question 8

Question 8 0.415 g of a pure acid, H₂X, is added to exactly 100 mL of 0.105 M NaOH(aq). A reaction occurs according to the equation H₂X(s) + 2NaOH(aq) → Na₂X(aq... show full transcript

Worked Solution & Example Answer:Question 8 0.415 g of a pure acid, H₂X, is added to exactly 100 mL of 0.105 M NaOH(aq) - VCE - SSCE Chemistry - Question 8 - 2008 - Paper 1

Step 1

i. the amount, in mol, of NaOH that is added to the acid H₂X initially.

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Answer

To calculate the amount of NaOH added, use the formula:

$n = C imes V$

Where:

$C$ is the concentration (0.105 M)
$V$ is the volume (100 mL = 0.100 L)

Substituting the values in:

$n = 0.105 imes 0.100 = 0.0105 ext{ mol}$

Step 2

ii. the amount, in mol, of NaOH that reacts with the acid H₂X.

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Answer

First, calculate the amount of NaOH in excess using the volume of HCl required for neutralisation:

Calculate the moles of HCl:

$n(HCl) = C imes V = 0.197 imes 0.02521 = 0.00497 ext{ mol}$

The total moles of NaOH added is 0.0105 mol, and the moles that react with the acid H₂X is:

$n(NaOH)_{reacting} = n(NaOH)_{initially} - n(NaOH)_{excess}$

So:

$n(NaOH)_{reacting} = 0.0105 - 0.00497 = 0.00553 ext{ mol}$

Step 3

iii. the molar mass, in g mol⁻¹, of the acid H₂X.

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Answer

To find the molar mass, use the formula:

$M = \frac{m}{n}$

Where:

$m$ is the mass of the acid (0.415 g)
$n$ is the amount of NaOH reacting with H₂X (0.00553 mol)

Thus:

$M_{(H₂X)} = \frac{0.415}{0.00553} = 75.05 ext{ g mol}^{-1}$

Question 8 0.415 g of a pure acid, H₂X, is added to exactly 100 mL of 0.105 M NaOH(aq) - VCE - SSCE Chemistry - Question 8 - 2008 - Paper 1

Worked Solution & Example Answer:Question 8 0.415 g of a pure acid, H₂X, is added to exactly 100 mL of 0.105 M NaOH(aq) - VCE - SSCE Chemistry - Question 8 - 2008 - Paper 1

i. the amount, in mol, of NaOH that is added to the acid H₂X initially.

ii. the amount, in mol, of NaOH that reacts with the acid H₂X.

iii. the molar mass, in g mol⁻¹, of the acid H₂X.

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