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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 s... 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 moles of NaOH, we apply the formula:

$n = C \times V$

Where:

$n$ is the number of moles
$C$ is the concentration in moles per liter (M)
$V$ is the volume in liters (L)

We convert 15.0 mL to liters: $V = 15.0 ext{ mL} = 0.0150 ext{ L}$

The concentration of NaOH is 2.50 M, so:

$n_{NaOH} = 2.50 ext{ M} \times 0.0150 ext{ L} = 0.0375 ext{ moles}$

Step 2

Determine the moles of oxalic acid

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Answer

From the reaction of oxalic acid with NaOH, we know the stoichiometry shows that 1 mole of oxalic acid reacts with 2 moles of NaOH. 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 the oxalic acid solution

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Answer

Now we can calculate the concentration of the oxalic acid using the formula:

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

Where:

$C$ is the concentration in moles per liter (M)
$n$ is the number of moles (0.01875 moles)
$V$ is the volume in liters (0.025 L)

Thus:

$C_{H_2C_2O_4} = \frac{0.01875 ext{ moles}}{0.025 ext{ L}} = 0.750 ext{ M}$

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