Titration curves

If I were to model the titration of $30 \ \text{cm}^{3}$ of $ 3 \ \text{mol dm}^{-3}$ $\ce{CH3COOH\_{(aq)}}$ with $3 \ \text{mol dm}^{-3}$ $\ce{NaOH\_{(aq)}} $ , what would the equation look like (graphing volume of base added vs. pH of the resulting system)? I am able to obtain a graph that looks like a titration curve by using the Henderson-Hasselbalch equation [more details below], but according to [Wikipedia](https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation), the equation fails when


* a strong acid or base is involved
* strong concentrations $> 1 \hspace{1mm} \text{mol dm}^{-3}$ are involved.


These conditions apply to this situation. Nevertheless, the equation I have found (it's very slightly not continuous) is 


$$f(v) =
\begin{cases}
4.76 + \log\_{10}{\left(\dfrac{3v}{0.09 - 3v} \right)}, & v \in ( 0 , 0.03) \\
8.86 & v = 0.03 \\
14 - \log\_{10}{\left(\dfrac{3v - 3}{0.03 - v} \right)}, & v \in (0.03 , \infty)
\end{cases}$$


I would like to know how to fix this issue (if there is any way to better represent this titration). Thank you.




---


**Before the equivalence point**


$$\text{moles of $\ce{H+}$}: 30 \ \text{cm}^{3} \cdot {1 \ \text{dm}^{3} \over 1000 \ \text{cm}^3} \cdot 3 \ \text{mol dm}^{-3} = 0.09 \ \text{mol} $$


$$\text{moles of $\ce{OH-}$}: v \cdot 3 \ \text{mol dm}^{-3} = 3v \ \text{mol}$$


*calculating the acid and its conjugate base*


using an ICE chart,


$$\ce{CH3COOH + OH- <=>CH3COO- + H2O}$$


$$\text{moles of $\ce{CH3COO-}$}: 3v \ \text{mol}$$


$$\text{moles of $\ce{CH3COOH}$}: 0.09 - 3v \ \text{mol}$$


$$[\ce{CH3COO-}] = {3v \over 0.09 + v} \ \text{mol dm}^{-3}$$


$$[\ce{CH3COOH}] = {0.09 - 3v \over 0.09 + v} \ \text{mol dm}^{-3}$$


*using Hendersen-Hasselbalch*


if $f(v)$ is pH,


$$f (v) = 4.76 + \log\_{10}{\left(\dfrac{3v}{0.09 - 3v} \right)}$$


**At the equivalence point**


*using an ICE chart*,


$$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$


$$K\_{b} = 10^{-9.24} = {v^{2} \over 0.09 - v}$$


$$0 = v^2 + 10^{-0.24} x - 0.09 \left(10^{-9.24} \right)$$


$$v = 8.86$$


**After the equivalence point**


To be honest, I think just screwed around with this graph until I got a similar shape. I knew that that


$$\lim\_{v \to \infty} f (v) = 14$$


at standard conditions, so I subtracted the log function from 14.


