When 25 mL of 0.30 M ammonium iodide and 25 mL of 0.30 M sodium hydroxide are combined, the pH of the resulting solution will be : A. Less than 7 B. Equal to 7 C. Greater than 7

pH of the resulting solution is greater than 7

Explanation:

Concentration of ammonium iodide = 25 x 0.3 mmol

                                                            = 7.5 mmol

And concentration of Sodium hydroxide solution = 25 x 0.3 mmol

                                                                                  = 7.5 mmol

Since the concentration of both solution is equal.

The  ionization reaction is

                               NH₄I + NaOH  →  NaI + NH₄OH

Concentration of NH₄OH is equal to 7.5 mmol because NH₄OH is the product of above reaction and the two electrolytes are strong, So completely ionized in solution.

so pOH = -log(7.5 x 10⁻³)

             = 2.12

We know pH + pOH = 14

               pH = 14 - 2.12 = 11.88

Since pH of the solution is greater than 7

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.