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Pani-rosa [81]
2 years ago
8

In general, the higher the charge on the ions in an ionic compound, the more favorable is the lattice energy. Why do some stable

ionic compounds have +1 charged ions even though +4, +5, and +6 charged ions would have a more favorable lattice energy?
Chemistry
1 answer:
Sever21 [200]2 years ago
7 0

Answer:

The main reason why some stable ionic compounds have +1 charged ions, even though other higher would have a more favorable lattice energy is because;

1. Energy is required to remove each of the electrons in an atom, and the more more the electrons are removed, the smaller the atom radius becomes, with increasing nuclear attraction between the opposite charges.

Note, it is a bit easy to remove a single electron, when compared to 4, 5 or 6 electrons. hence, the reason why Ionic compounds are more stable with + charged ions than others.

2. The stability of an ionic compound is dependent on his lattice energy. And the ions with large charge will have a greater attraction between the cations and anions, hence have higher lattice energy. However, the distance between the ions and the nucleus of the atom plays a major role in the stability of the ion.

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Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

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The pK_a value is pK_a  =7.82

Explanation:

From the question we are told

    The volume of base is  V_B = 26.mL = 0.0260L

     The pH of solution is  pH =  7.82

      The concentration of the acid is C_A = 0.1M

From the pH we can see that the titration is between a strong base and a weak acid

 Let assume that the the volume of acid is  V_A = 18mL= 0.018L

Generally the concentration of base

                    C_B = \frac{C_AV_A}{C_B}

Substituting value  

                     C_B = \frac{0.1 * 0.01800}{0.0260}

                    C_B= 0.0692M

When 13mL of the base is added a buffer is formed

The chemical equation of the reaction is

           HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}

Now before the reaction the number of mole of base is  

            No \ of \ moles[N_B]  =  C_B * V_B

Substituting value  

                    N_B = 0.01300 * 0.0692

                         = 0.0009 \ moles    

                                 

Now before the reaction the number of mole of acid is  

            No \ of \ moles  =  C_B * V_B

Substituting value  

                    N_A = 0.01800 *0.1

                         = 0.001800 \ moles

Now after the reaction the number of moles of  base is  zero  i.e has been used up

    this mathematically represented as

                         N_B ' = N_B - N_B = 0

    The  number of moles of acid is  

             N_A ' = N_A  - N_B

                   = 0.0009\ moles

The pH of this reaction can be mathematically represented as

                 pH  = pK_a + log \frac{[base]}{[acid]}

Substituting values

                  7.82 = pK_a +log \frac{0.0009}{0.0009}

                  pK_a  =7.82        

                     

             

                                 

       

           

                     

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