Lecture Notes – Ch 15.5 & 15.6      Std 5

 

 

Ch 15.5 – pH of strong Acids

           

         For the preclass today we did these problems:

            Page 506, problem 53 in class and you will see problems like this on the quiz.  Here are the answers:

            53.  a.  [H+]  =  anti-log [-8.51]  =  3.1 x 10-9 M

                        pOH = 14 - 8.51  =  5.49

                        [OH-]  =  anit-log [-5.49]  =  3.2 x 10-6 M

                  b.   [0H-]  =  anti-log [-9.39]  =  4.1 x 10-10 M

                        pOH = 14 – 9.39  =  4.61

                        [H+]  =  anit-log [-4.61]  =  2.5 x 10-5 M

                  c.   [H+]  =  anti-log [-2.54]  =  2.9 x 10-3 M

                        pOH = 14 – 2.54  =  11.46

                        [OH-]  =  anit-log [-11.46]  =  3.5 x 10-12 M

                  d.   [0H-]  =  anti-log [-4.82]  =  1.5 x 10-5 M

                        pOH = 14 – 4.82  =  9.18

                        [H+]  =  anit-log [-9.18]  =  6.6 x 10-10 M

 

         FOR STRONG ACIDS

 

            Concentrations of [H+] ions = concentrations given of strong acids

 

            Example:

 

         HCl  ŕ  H+  +  Cl-      Notice !!!   No backwards arrow for the equilibrium.  This means that the HCl ionizes 100% in water. 

 

         Next we did problem 57 on page 506, from the homework.  As you see in each problem, the [H+] is always the same as the concentration of the strong acid.  That is because all the HCl become H+ and Cl- ions.  Here are the answers:

               P. 506   problem 57  

               a.         [H+]    1.04 x 10-4 M and to find the pH we take the - log [H+]  =  3.98

               b.         [H+]     0.00301 or 3.01 x 10-3 M and to find the pH we take the - log [H+]  =  2.5

               c.         [H+]    5.41 x 10-4 M and to find the pH we take the - log [H+]  =  3.27

               d.         [H+]    6.42 x 10-2 M and to find the pH we take the - log [H+]  =  1.19

           

Ch 15.6 – Buffers

 

         Buffers are created by mixing a weak acid and the salt of its conjugate base.

                       

HC2H3O2  <------------------------  H+   +   C2H3O2-  

(weak acid)              -->

 

         Notice the equilibrium favors the acid (left) very strongly.  That is why it is a weak acid.  Not many ions get formed because it doesn’t ionize much.

           

NaC2H3O2  -------------------------->  Na+  +  C2H3O2-

(salt of conj. base)

 

         Notice the salt ( NaC2H3O2 ) ionization is 100% , which means that all of the salt is ionized and none of the original NaC2H3O2 is left.  This puts a lot of the Na+ and C2H3O2-  ions into the water.

 

         After we put   HC2H3O2    and   NaC2H3O2  into the water, the ions and substances that can affect the pH are:

 

                                    HC2H3O2     and     C2H3O2-      (IN THE WATER)

 

         We now call this solution a buffer.  This buffer will have a pH = 4 (±2) depending on the concentrations of the ions. 

 

         Let’s see what happens when we add acids or bases to the buffer:

        

a.  ADD ACID 

HNO3   ----->   H+  +  NO3-       NEXT       H+  +  C2H3O2-  ----->  HC2H3O2

(strong acid)                             IT MIXES & REACTS

 

The  H+  ions from the HNO3 make the solution more acidic (lower pH), but they are quickly   “grabbed”  by the C2H3O2- ions which act as “assassin” ions and and hold on to the H+ ions.  This will  form HC2H3O2, which returns the pH to where is was.  Because the H+ isn’t in the water any more, the [H+]  returns to where is was before we put in the HNO3, which is pH = 4 .

 

         b.  ADD BASE

NaOH  ---->  Na+  +  OH-      NEXT    OH-  +  HC2H3O2  ---->  H2O  +  HC2H3O2

(strong base)                          IT MIXES & REACTS

 

The OH-  ions act like “assassins” and  “grab”  the  H+  awau from the HC2H3O2  to form water.  That  takes the OH- ions out of the water and makes more C2H3O2- .   Adding the OH-  will at first make the solution more basic for a short time, but when this second reaction takes place the pH returns to pH = 4  because the OH- ions are used up forming water.

 

Here are additional explanations of Buffers given in class just before Quiz 3.1

 

         Buffer examples

 

            1.      Hydrofluoric acid (HF) a weak acid,  and potassium fluoride (KF) the salt of the conjugate base.

           

HF                                    -->                   H+    +     F-

Weak acid    <------------------------                    conjugate base

 

             KF    ------------------------>    K+    +    F-

salt of  conj base

 

         If we add OH-  (NaOH)  -  What takes out the OH-  so the pH doesn’t change much?  

 

         If we add H+ (HCl) – What takes out the H+ so the pH doesn’t change much?

        

                                                            Answers

 

         2.     Acetic acid (HC2H3O2 ), and sodium acetate (NaC2H3O2 ), the salt of the conjugate base.

        

HC2H3O2                          -->           H+     +    C2H3O2-

Weak acid   <----------------------                  conjugate base

 

NaC2H3O2    ---------------------->  Na+    +   C2H3O2-

salt of conjugate base

 

 

         If we add OH-  (NaOH)  -  What takes out the OH-  so the pH doesn’t change much?        If we add OH-  (NaOH)  -  What takes out the OH-  so the pH doesn’t change much?  

 

         If we add H+ (HCl) – What takes out the H+ so the pH doesn’t change much?

 

 

         If we add H+ (HCl) – What takes out the H+ so the pH doesn’t change much?

 

                                                Answers

 

3.   Can salts of strong acids act as buffers by themselves?

         Examples:     NaCl,  KCl, CaCl2 , MgCl2                Answers

 

 

4.  Here is an important summary of buffer information:

     The a  and  b below explain how a buffer works to keep the pH relatively constant when acids or bases are added to their solutions.

 

         a.  Buffers resist change in pH even if H+  or  OH-  ions are added from strong acids or bases.

 

b.  H+ ions react with conjugate bases of weak acids.  OH- ions react with the H+ ions from weak acid and take the hydrogen away from them.