What is the ph of a buffer prepared by adding 0.506 mol of the weak acid ha to 0.305 mol of naa in 2.00 l of solution? the dissociation constant ka of ha is 5.66*10^-7.ph = is the \rm ph after 0.150 mol of \rm hcl is added to the buffer from part a? assume no volumechange on the addition of the acid. ph = is the \rm ph after 0.195 mol of \rm naoh is added to the buffer from part a? assume novolume change on the addition of the base. ph =

A. pH= 6.028

b. pH = 5.620

C .pH =  6.453

Explanation:

A buffer's pH is easily calculated by using the Henderson-Hasselbalch equation which relates the pH of the buffer to the  p K a  of the weak acid in the buffer and the concentrations of the acid and its conjugate base in the buffer. It is mathematically expressed as:

p H = p K a + l o g[ A − ]  / [ H A ]

First we will calculate the  p K a  of the weak acid from the given  K a , as well as the concentration of HA and  A − :

p K a = − log K a = − log ( 5.66 × 10 − 7 ) = 6.247  m o l e

H A  =0.506m o l e HA  / 2 L = 0.253M

[ A − ] = 0.305 m o l e A  / 2.00  L = 0.153 M

Now we will use the Henderson-Hasselbalch equation to calculate the pH

pH =  6.247 + log ( 0.153 / 0.253)

      =  6.247-0.21842909035

       = 6.028

Part B

Because  A −  react with HCl in 1-to-1 molar ratio and HCl is the limiting reactant, all of the HCl added will react with completely with the  A − in the buffer to form HA. The new concentration of HA and  A −  are:

[ H A ] = 0.150+0.506 / 2 = 0.328M

{A}   = ( 0.305 − 0.150 ) m o l e A  /  2.00 L = 0.0775 M

Applying the Henderson-Hasselbalch equation again

p H  = 6.247 + log ( 0.0775 / o.328) = 5.620

PART C:

NaOH reacts with HA in 1-to-1 molar ratio and NaOH is the limiting reactant in this case, all the NaOH will react with some of the HA to form more  

[ A − ]  in the buffer. The new concentrations of HA and  [ A − ]  are:

[ H A ]  = 0.506-0.195 / 2 = 0.1555 M

[A-]   = 0.305 +0.195 / 2 = 0.250M

Applying the Henderson-Hasselbalch equation again

p H  = 6.247 + log 0.250 / 0.1555

pH =  6.453