# 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, 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.