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Gwar [14]
2 years ago
15

Consider the following system at equilibrium. CaCO3(s) Double headed arrow. Ca2 (aq) CO32â€""(aq) The addition of which compound

will cause a shift in equilibrium because of a common ion effect? CCl4 CO2 CuSO4 Na2CO3.
Chemistry
1 answer:
Dominik [7]2 years ago
6 0

The compound that has been capable to shift the equilibrium with common ion effect has been  \rm \bold{Na_2CO_3}.  Thus, option D is correct.

The complete equation for the decomposition of calcium carbonate has been:

\rm CaCO_3\rightleftharpoons Ca^2^+\;+\;CO_3^2^-

The equilibrium has been the state in the reaction, when the number of products and has been equivalent to the number of reactants. The increase or decrease in the concentration of any of the reactant or product results in the shift in the equilibrium.

For the given reaction, the addition of Calcium or carbonate ions to the solution results in the increase in the product concentration with the common ion effect, and thereby shifts the equilibrium condition.

The ions liberated by the following compounds has been:

\rm CCl_4\;=\text{not dissolved}\\CO_2\;+\;H_2O\leftrightharpoons CO_3^-\;+\;H^+\\CuSO_4\leftrightharpoons Cu^2^+\;+\;SO_4^2^-\\Na_2CO_3\leftrightharpoons 2\;Na^+\;+\;CO_3^-

The compound, that has been capable of liberating the carbonate ions in the medium has been \rm \bold{Na_2CO_3}. Thus, it results in the shift of the equilibrium due to common ion effect. Hence, option D is correct.

For more information about common ion effect, refer to the link:

brainly.com/question/4090548

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3 years ago
What is the Phorgum?
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7 0
4 years ago
If you only have 36.5 grams of Potassium Carbonate (K2CO3 molar mass = 138.2 g/mole). How many liters of a 0.52 M solution can y
kirza4 [7]

Answer:

0.508 L of solution.

Explanation:

Always a safe bet to convert to moles:

n= {m \over MM}\\

Where n is moles, m is mass, and MM is molar mass.

Now remember the equation for concentration (molarity):

C={n \over V}

Where C is the concentration, n is moles, and V is volume.

To make this easy, combine the two equations (note n appears in both, so you can do a substitution) and solve for V as the question asks:

C={m \over MM \times V}\\0.52={36.5 \over 138.2 \times V}\\V={36.5 \over 138.2 \times 0.52}\\V=0.508

Therefore we can make 0.508 L of solution.

8 0
3 years ago
A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles
alexgriva [62]

Answer : The pH of the solution is, 3.41

Explanation :

First we have to calculate the moles of HF.

\text{Moles of HF}=\text{Concentration of HF}\times \text{Volume of solution}

\text{Moles of HF}=0.250M\times 1.50L=0.375mol

Now we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (6.8\times 10^{-4})

pK_a=4-\log (6.8)

pK_a=3.17

The reaction will be:

                             HF+OH^-\rightleftharpoons F^-+H_2O

Initial moles     0.375     0.100   0.375

At eqm.   (0.375-0.100)      0     (0.375+0.100)

                     = 0.275                    = 0.475

Now we have to calculate the pH of solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[F^-]}{[HF]}

Now put all the given values in this expression, we get:

pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]

pH=3.41

Thus, the pH of the solution is, 3.41

8 0
3 years ago
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