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Vladimir79 [104]
3 years ago
8

Which pair of elements can form ionic bonds?

Chemistry
2 answers:
Oxana [17]3 years ago
7 0

Answer:

Correct answer: E.  potassium (K) and bromine (Br)

Explanation:

(Correct answer for (portal and Edmentum users too)

Hope I was helpful :) good luck on your school work and have a great day

Sindrei [870]3 years ago
3 0

Answer:

E.  potassium (K) and bromine (Br)

Explanation:

An ionic bond is formed between compounds with a large electronegativity difference between them. It is usually between a metal and non-metal.

  • Potassium is a true metal found in group 1 on the periodic table.
  • Bromine is a highly electronegative non-metal which is a halogen.
  • Potassium will lose one of its electrons which will be gained by the Bromine.
  • The electrostatic attraction between the two species will cause the ionic bond to form.
  • The ability of one specie willing to lose electron and the other gaining, is the main bed rock of ionic bonding.
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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 7.8
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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?

Answer:

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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