A 75.0-mLmL volume of 0.200 MM NH3NH3 (Kb=1.8×10−5Kb=1.8×10−5) is titrated with 0.500 MM HNO3HNO3. Calculate the pHpH after the addition of 13.0 mLmL of HNO3

The pH is  

Explanation:

From the question we are told that

         The volume of   is  

         The concentration of is  

          The concentration of is

          The volume of    added  is

           The base dissociation constant is

The number of moles of  that was titrated can be mathematically represented as

substituting values

The number of moles of that was titrated can be mathematically represented as

substituting values

So from the calculation above the limited reactant is  

The chemical equation for this reaction is

From the chemical reaction

  1 mole of  is  titrated with 1 mole of to produce 1 mole of  NH^{4+}

So

    0.0065  moles of  is  titrated with 0.0065 mole of to produce 0.0065 mole of  

So

  The remaining  moles of  after the titration is

=>        

Now according to Henderson-Hasselbalch equation the pH of the reaction is mathematically represented as

Where is mathematically represented as

Now        

Where is the ionization constant  of   with value

Hence    

Substituting this into the equation

substituting values

c is the answer because when something is left after something evaporates that shows the chemical change.