[larrysz92]

Here is the question:

A buffer is created by sisolving 0.45 mol of sodium ascorbate (NaC6H7O6) and 0.50 mol ascorbic acid (HC6H7O6) in enough water to create 1.00 L of solution. THe Ka of ascorbic acid is 7.9 x 10^-5

a) What is the original pH of the buffer?
b) What is the pH when 0.015 mol of HCl is added to the buffer?
c) What is the pH when 1.00g of solid NaOH is added to the buffer?

This is what i have done

a)   HA(aq) + H2O(l) <-> H3O+(aq) + A-(aq)

Ka=[H3O+] [A-]
             [HA]

ICE   [HA] in mol/L   [H3O+] in mol/L   [A-] in mol/L
Initial   0.45   0   0.450
Change   -x   +x   +x
Equilibrium   0.5 - x   x   0.45 + x
Revised Equilibrium   0.5   x   0.45

cHA=n/V                       cA =n/V
      =0.5/1                        =0.45/1
      =0.5 mol/L        =0.45 mol/L

Applying 100 Rule:
[Original]/Ka = 0.5/7.9 x 10-5 = 6329>100   

Ka=[H3O+] [A-]
             [HA]

7.9 x 10-5= x (0.45)
                  0.5
x = 7.9 x 10-5 (0.5)
         0.45
x = 8.8 x 10-5 mol/L

[H3O+] = 8.8 x 10-5 mol/L

pH = -log[H3O+] = -log(8.8 x 10-5) = 4.0556 = 4.06 (3 Sig Fig)

b)

ICE   [HA] in mol/L   [H3O1+]in mol/L       [A1-]  in mol/L
Initial   0.515            0                             0.435
Change   -x                     +x                              +x
Equilibrium    0.515-x      x                          0.435+x
Revised Equilibrium   0.515   x                       0.435

HCL is a strong acid and will cause the reaction to shift to the right. The [A^(1-)] will be reduced to 0.45 mol/L-0.015 mol/L=0.435 mol/L. The [HA] will increase to 0.50 mol/L+0.015 mol/L=0.515 mol/L.
Applying the 100 rule will yield the same result so it is redundant to do the calculation again.
7.9×〖10〗^(-5)=x(0.435)/0.515
x=(7.9×〖10〗^(-5) (0.515))/0.435
x=9.4×〖10〗^(-5)
[H_3 O]=9.4×〖10〗^(-5)  mol/L
pH=-log⁡[9.4×〖10〗^(-5) ]=4.02687=4.0(rounded for significant figures)
Therefore, after adding the HCL the pH of the buffer is 4.0

c) Molar mass of NaOH=22.99+15.99+1.01=39.99
n=(1.0 g)/(39.99 g/mol)=2.5×〖10〗^(-2)  mol
The 2.5×〖10〗^(-2)  mol added to the [A^(1-)] will increase the concentration to 0.45 mol/L+2.5×〖10〗^(-2) mol/L=0.475 mol/L. The equilibrium will shift to the right and reduce the amount of [HA] to 0.50 mol/L+2.5×〖10〗^(-2) mol/L=0.475mol/L.
ICE   [HA] in mol/L   [H_3 O^(1+) ]in mol/L   [A^(1-) ]  in mol/L
Initial   0.475   0   0.475
Change   -x   +x   +x
Equilibrium    0.475-x   x   0.475+x
Revised Equilibrium   0.475   x   0.475

7.9×〖10〗^(-5)=(x(0.475))/0.475
x=(7.9×〖10〗^(-5) (0.475))/0.475
x=7.9×〖10〗^(-5)
[H_3 O]=7.9×〖10〗^(-5)
pH=-log⁡(7.9×〖10〗^(-5) )=4.1023=4.1(rounding for significant figures)
Therefore, after adding the NaOH the pH of the buffer is 4.1


I dont think i did them right, can someone help me pls?

Thanks

[Borek]

Why don't you use Henderson-Hasselbalch equation?