What is the pH at both equivalence points of titration between diprotic tartaric acid and NaOH? Here's all the info I have and my attempt in solving the question:

The chemical equation for the titration:

$$\ce{C4H4O6 + 2NaOH -&gt; Na2C4H4O6 + 2H2O}$$

Here are the 1st and 2nd $\mathrm{Ka}$ value for tartaric acid:

$$K_\mathrm{a1} = 9.2 \times 10^{-4}, K_\mathrm{a2} = 4.3 \times 10^{-5}$$

Molarity of tartaric acid = $5 \times 10^-4$

Standard solution of $\ce{NaOH }$ was made using $\pu{0.98 g}$ of solid $\ce{NaOH}$ dissolved in $\pu{250 ml}$ of water. The molar mass of $\ce{NaOH}$ is $\pu{39.997 g/mol}$.

This is what I have done so far.

n (C4H4O6) = cv = (5 × 10-4 M) (25ml ÷ 1000L) = 0.02 mol

n (NaOH) = m ÷ M = (0.98g) ÷ (39.997 g/mol) = 0.0245 mol (4dp).

C (NaOH) = n ÷ v = (0.0245 mol) ÷ (250ml ÷ 1000L) = 0.098M

N (NaOH) at equivalence point = cv = (0.098M) (0.204L) = 0.02 mol

C4H4O6(aq)+ 2NaOH(aq) → Na2C4H4O6(aq) + 2H2O(l)

I 0.02         0.02           0            -
C - 0.02      - 0.02       + 0.02          -
E   0           0            0.02          -

C (Na2C4H4O6) = n ÷ v = 0.02 ÷ (0.025L + 0.25L) = 0.0727M (4dp)

Na2C4H4O6(aq) + H2O(l) → C4H4O4Na2^+(aq) + OH^-(aq) below is ice table, stackexchange won't allow me to seperate the values.

I  0.0727          -           0              0
C   - x            -         + x             + x
E 0.0727 – x       -          x               x

Kb = Kw ÷ Ka = (1 × 10-14) ÷ (9.2 × 10-4) = 1.087 × 10-11

Kb = [C4H4O4Na2+] [OH-] ÷ [Na2C4H4O6] 1.087 × 10-11 = x2 ÷ 0.0727 x = √ ((1.087 × 10-11) (0.0727)) x = 8.89 × 10-7

pOH = - log [OH-] = - log (8.89 × 10-7) = 1.9

pH = 14 – pOH = 14 – 1.9 = 12.1
For the 1st equivalence point, the tartaric acid $\ce{H2A}$ is titrated to $\ce{NaHA}$

For the solutions of acid salts, we can with advantage use the formula

$$\mathrm{pH}=\frac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{2}$$

that can be derived from equilibrium

$$\ce{2 HA- &lt;=&gt; H2A + A^2-}$$

For the 2nd equivalence point there is  $\ce{Na2A}$ solution, which is hydrolyzed with equilibrium

$$\ce{A^2- + H2O &lt;=&gt; HA- + OH-}$$

With neglecting OH- from water hydrolysis, $[\ce{HA^-}] = [\ce{OH^-}]$

$$\begin{align}
K_\mathrm{a2}&amp;=\frac{[\ce{H+}][\ce{A^2-}]}{[\ce{HA-}]}\\
\mathrm{p}K_\mathrm{a2}&amp;=\mathrm{pH} - \log{[\ce{A^2-}]} +\log {[\ce{HA-}]}\\
&amp;=\mathrm{pH} - \log{[\ce{A^2-}]} +\log {[\ce{OH-}]}\\
&amp;=\mathrm{pH} - \log{[\ce{A^2-}]} -\mathrm{pOH}\\
&amp;=\mathrm{pH} - \log{[\ce{A^2-}]} -14 + \mathrm{pH}\\
\mathrm{pH}&amp;=7+\frac12\left(\mathrm{p}K_\mathrm{a2} +\log{ c_\mathrm{\ce{Na2A}}}\right)\\
\end{align}$$

The molarity of $\ce{NaOH}$ solution $c_{\ce{NaOH}}=\frac{m}{M\times V}=\pu{0.09801  mol/L}$

The volume of $\ce{NaOH}$ solution for given volume of tartaric acid solution

$$V_{\ce{NaOH}}=V_{\ce{H2A}}\times \frac{2\times c_{\ce{H2A}}}{c_{\ce{NaOH}}}\\=0.0102\times V_{\ce{H2A}}$$

The final concentration 

$$\begin{align}
[\ce{Na2A}]&amp;=\frac{[\ce{H2A_\mathrm{init}}]\cdot V_{\ce{H2A}}}{V_{\ce{H2A}} +V_{\ce{NaOH}}}\\
&amp;=\frac{[\ce{H2A_\mathrm{init}}]\cdot V_{\ce{H2A}}}{V_{\ce{H2A}} + 
0.0102  \cdot  V_{\ce{H2A}}}\\
&amp;=\frac{[\ce{H2A_\mathrm{init}}]}{1+ 0.0102}\\
&amp;=5\times 10^{-4}/1.0102\\
&amp;=\pu{4.9495\times10^{-4} mol/L}\\
&amp;\overset{\mathrm{rounding}}=\pu{4.95\times 10^{-4} mol/L}\\
\end{align}$$