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4 years ago

How many ionization constants are associated with oxalic acid (H2C2O4)?

Chemistry

2 answers:

Bas_tet [7]4 years ago

6 0

There are 2 ionization constants associated with oxalic acid.

victus00 [196]4 years ago

3 0

Answer: Two.

Explanation: Number of ionization constants associated with an acid equals to the number of hydrogen ions(protons) that it produces in aqueous solution.

For example, HCl is a monoprotic acid as it produces only one hydrogen ion in aqueous solution and so only one ionization constant is associated with it.

$H_2SO_4$ (sulfuric acid) is a diprotic acid as it produces two hydrogen ions in aqueous solution and hence there are two ionization constant values associated with this.

Similarly, Oxalic acid is also a diprotic acid and so two ionization constant values are associated with this.

The two ionization equations of oxalic acid are as follows:-

$H_2C_2O_4\rightleftharpoons H^++HC_2O_4^-$

$HC_2O_4^-\rightleftharpoons H^++C_2O_4^-^2$

$Ka_1$ and $Ka_2$ are the two ionization constants for first and second equations respectively.

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The solubility of oxygen gas in water at 40 ∘c is 1.0 mmol/l of solution. What is this concentration in units of mole fraction?

juin [17]

The formula for mole fraction is:

$mole fraction of solute = \frac{number of moles of solute}{total number of moles of solution}$ -(1)

The solubility of oxygen gas = 1.0 mmol/L (given)

1.0 mmol/L means 1.0 mmol are present in 1 L.

Converting mmol to mol:

$1.00 mmol\times \frac{1 mol}{1000 mmol} = 0.001 mol$

So, moles of oxygen = 0.001 mol

For moles of water:

1 L of water = 1000 mL of water

Since, the density of water is 1.0 g/mL.

$Density = \frac{mass}{volume}$

$Mass = 1.0 g/ml\times 1000 mL = 1000 g$

So, the mass of water is 1000 g.

Molar mass of water = 18 g/mol.

Number of moles of water = $\frac{1000 g}{18 g/mol} = 55.55 mol$

Substituting the values in formula (1):

$mole fraction = \frac{0.001}{55.55+0.001}$

$mole fraction = 1.8\times 10^{-5}$

Hence, the mole fraction is $1.8\times 10^{-5}$ .

7 0

3 years ago

A chemist must dilute of aqueous aluminum chloride solution until the concentration falls to . He'll do this by adding distilled

marshall27 [118]

Answer:

0.257 L

Explanation:

The values missing in the question has been assumed with common sense so that the concept could be applied

Initial volume of the AICI3 solution $=23.1 \mathrm{mL}$

Initial Molarity of the solution $=833 \mathrm{mM}$

Final molarity of the solution $=75.0 \mathrm{mM}$

Final volume of the solution $=?$

From Law of Dilution, $M_{f} V_{f}=M_{i} V_{i}$

$\Rightarrow V_{f}=\frac{M_{i} V_{i}}{M_{f}}=\frac{833 \mathrm{mM} \times 23.1 \mathrm{mL}}{75.0 \mathrm{mM}}=256.564 \mathrm{mL}=0.256564 \mathrm{L}=0.257 L$