Consider a titration involving a weak acid, HA (Ka = 7.89x10-6). A standardized solution NaOH(aq) with a concentration of 0.341M is used as?

a)
HA(aq) + NaOH(aq) → NaA(aq) + H₂O(ℓ)
Mole ratio  HA : NaOH = 1 : 1

Moles of NaOH reacted = (0.341 mol/L) × (16.73/1000 L) = 0.005705 mol
Moles of HA required = 0.005705 mol
Concentration of HA = (0.005705 mol) / (20.00/1000 L) = 0.285 M

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b)
Consider the dissociation of HA:
                HA(aq) + H₂O(ℓ) ⇌ H₃O⁺(aq) + A⁻(aq)    Ka = 7.89 × 10⁻⁶
Initial:      0.285 M                   0 M            0 M
Change:    -y M                      +y M         +y M
Eqm:    (0.285 - y) M               y M            y M
             ≈ 0.285 M

At eqm:
Ka = [H₃O⁺] [A⁻] / [HA]
y² / 0.285 = 7.89 × 10⁻⁶
y = √(0.285 × 7.89 × 10⁻⁶) = 1.50 × 10⁻³
pH = -log[H₃O⁺] = -log(1.50 × 10⁻³) = 2.82

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c)
HA(aq) + NaOH(aq) → NaA(aq) + H₂O(ℓ)
Mole ratio  NaOH : NaA = 1 : 1

Moles of NaOH reacted = 0.005705 mol
Moles of NaA produced = (16.73 + 20.00) mL = 36.73 mL = 0.03673 L
Concentration of HA in the final soluton = (0.005705 mol) / (0.03673 L) = 0.155 M
[A⁻] in the final solution = 0.155 M

Consider the dissociation of A⁻:
                 A⁻(aq) + H₂O(ℓ) ⇌ HA(aq) + OH⁻(aq)    Kb = (1.00 × 10⁻¹⁴)/(7.89 × 10⁻⁶)
Initial:      0.155 M                    0 M         0 M
Change:    -z M                      +z M        +z M
Eqm:     (0.155 - z) M               z M         z M
              ≈ 0.155 M

At eqm:
Ka = [H₃O⁺] [A⁻] / [HA]
y² / 0.155 = (1.00 × 10⁻¹⁴)/(7.89 × 10⁻⁶)
y = √[0.155 × (1.00 × 10⁻¹⁴)/(7.89 × 10⁻⁶)] = 1.40 × 10⁻⁵
pH = 14 + log[OH⁻] = 14 + log(1.40 × 10⁻⁵) = 9.15

HA + NaOH = H2O +NaA

moles of NaOH at the equivalence point =  0.341 moles/ liter *0.01673 L 
=  5.70^ 10 ^-3 moles of NaOH
according to our equation these are the moles of weak acid at the equivalence point 

these are in 20.00 m so molarity = 5.70 * 10 ^-3 moles/ 0.02000L 
= 0.285 M weak acid 

[H] initially  =  SQRT(Ka*C) = SQRT(7.89x10-6 * 0.285 ) = 0.00150 M
pH = 2.82 initially 

at the equivalence point we have :

total volume = 36.73 mL so 
[HA] =  at the equivalence point 5.70 * 10 ^-3 moles / .03673 = 0.1552M

however some A - will react with the water to give an alkaline solution 
pH ~ 8.2