Question

Calculate the number of moles of HCl(g) that must be added to 1.0 L of 0.80 M NaC2H3O2 to produce a solution buffered at pH = pKa. Ka(CH3CO2H) = 1.8×10-5 Number of moles = mol Submit b Calculate the number of moles of HCl(g) that must be added to 1.0 L of 0.80 M NaC2H3O2 to produce a solution buffered at pH = 4.20. Ka(CH3CO2H) = 1.8×10-5 Number of moles = mol c Calculate the number of moles of HCl(g) that must be added to 1.0 L of 0.80 M NaC2H3O2 to produce a solution buffered at pH = 5.00. Ka(CH3CO2H) = 1.8×10-5 Number of moles = mol

Answer 1

Answer:

a.  0.40

b.  0 .62 mol HCl

c.   0.29 mol HCl

Explanation:

Lets call A⁻ the molar concentration of the weak base A⁻, and HA the acid concentration of HA

a. pH = pKa

from Henderson-Hasselbach equation for buffer solutions:

pH = pKa - log(( A⁻/HA))

we can see that pH = Pka when the  log(( A⁻/HA)) is 0 which is the case for the log 1 hence

A⁻ =HA  and

Ka=  H⁺ x  A ⁻/ HA  

Now for the equilibrium of the weak acid  we have

R                     HA==================  H⁺ + A⁻

Initially            0.80                                    0     0

Change            -x                                     + x    +  x

Equilibrium     0.80 - x                               x      x

Since x = A⁻ and HA = 0.80- x

x/ (0.80-x) =1  

x = 0.80 -x      2x = 0.80

x = 0.40

b. From the Henderson-Hasselbach eqn:

pH = pKa - log ((A⁻/HA))

4.20-4.74 = log ((0 .80-x)/x)  (the negative log of A⁻/HA is the same as log HA/A⁻)

-0.54 = log ((0.80-x)/x)

0.29 = (0.80-x)/x

0.29 x = 0.80 - x

1.29 x = 0.80 moles HCL

x = 0.62

c.  pH= 5.00

Using the same relationships as in b.

5.00-4.74 = log ((0.80-x)/x)

1.80 =( 0.80 - x) /x

2.8 x = 0.80

x = 0.29

0.29 mol HCl needed