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The structure of oxalic acid is shown below - VCE - SSCE Chemistry - Question 18 - 2010 - Paper 1

Question 18

The structure of oxalic acid is shown below. A 25.0 mL solution of oxalic acid reacts completely with 15.0 mL of 2.50 M NaOH. The concentration of the oxalic acid ... show full transcript

Worked Solution & Example Answer:The structure of oxalic acid is shown below - VCE - SSCE Chemistry - Question 18 - 2010 - Paper 1

Step 1

Calculate the moles of NaOH used

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Answer

To find the number of moles of NaOH, use the formula:

$n = C imes V$

Where:

$n$ = number of moles
$C$ = concentration (M)
$V$ = volume (L)

Here, the volume of NaOH is 15.0 mL, which is equivalent to 0.015 L:

$n_{NaOH} = 2.50 ext{ M} imes 0.015 ext{ L} = 0.0375 ext{ moles}$

Step 2

Determine the moles of oxalic acid

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Answer

The reaction between oxalic acid and NaOH is a 1:2 ratio since oxalic acid is a diprotic acid:

$ext{H}_2 ext{C}_2 ext{O}_4 + 2 ext{NaOH} \rightarrow ext{Na}_2 ext{C}_2 ext{O}_4 + 2 ext{H}_2 ext{O}$

Thus, the moles of oxalic acid are:

$n_{H_2C_2O_4} = \frac{n_{NaOH}}{2} = \frac{0.0375}{2} = 0.01875 ext{ moles}$

Step 3

Calculate the concentration of oxalic acid

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Answer

Now apply the formula for concentration:

$C = \frac{n}{V}$

Where:

$C$ = concentration (M)
$n$ = moles of solute
$V$ = volume of solution (L)

The volume of the oxalic acid solution is 25.0 mL or 0.025 L:

$C_{H_2C_2O_4} = \frac{0.01875}{0.025} = 0.75 ext{ M}$

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