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Nikitich [7]
3 years ago
6

Enter a balanced equation for the dissolution of caco3.

Chemistry
1 answer:
Tatiana [17]3 years ago
5 0

Answer:

CaCO3(aq) → Ca2+(aq ) + CO3 2-(aq)

Explanation:

  • Dissolution reactions are reactions that occur when a solute either in gaseous, liquid, or solid form dissolves in a solvent such as water to form a solution.
  • In this case we are given Calcium carbonate (CaCO3) which undergoes dissolution according to the equation;
  • CaCO3(aq) → Ca2+(aq ) + CO3 2-(aq)
  • Then<em><u> the bicarbonate ion combines with two protons from water to form a weak acid H2CO3. The weak acid is then broken down to form CO2 and H2O since its unstable.</u></em>
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Answer:- The Ka for the acid is 1.53*10^-^5 .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

Ka = [H^+][A^-]\frac{1}{HA}

Where, Ka is the acid ionization constant. Let's plug in the values.

Ka = \frac{X^2}{0.25-X}

Let's calculate the value of X first using the equation:

pH = -log[H^+][/tex]

on taking antilog ob above equation we get:

[H^+]=10^-^p^H

[H^+]=10^-^2^.^7^1

[H^+] = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

Ka=\frac{(0.00195)^2}{0.25-0.00195}

Ka=1.53*10^-^5

So, the value of Ka for butyric acid is 1.53*10^-^5 .

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