Below is a picture of a titration between a weak acid, $\ce{HA}$, and a 0.150 M $\ce{NaOH}$ solution. Find the initial volume of the weak acid in milliliters in this titration. enter image description here

My work so far: From the graph, I think that the two most important points are (0,3.0) and (20.00,8.0) - I don't know the reason, but it looks important.
At (0,3.0), there is no $\ce{NaOH}$. This equilibrium takes place: $$ \ce{HA_{(aq)} + H2O_{(l)} <=> A_{(aq)}- + H3O+_{(aq)} } $$

The constant $K_\text{a}$ can be found from: $$ K_\text{a} = \frac{[ \ce{H_3O^+}][\ce{A^-}]}{[\ce{HA}]} $$ with $[\ce{H_3O^+}] = 10^{-3}$ since $\text{pH}=-\log_{10}[\ce{H_3O^+}]=3$
This is a dead end for me.

At (20.00, 8.0), $[\ce{H_3O^+}] =10^{-8}$, this number has decreased because some $\ce{H_3O+}$ have reacted with the newly dissolved $\ce{OH^-}$, although how much I can't tell. I have reached another dead end

Other facts I do know:

  • 0.150 M of $\ce{NaOH}$ means for every 1 liter of the solution, there will be exactly 0.150 moles of $\ce{NaOH}$.
  • $\ce{NaOH}$ dissolves completely as a strong base. So, for every 1 mole of $\ce{NaOH}$ there will be 1 mole of $\ce{OH^-}$

2 Answers 2



There are three main points of interest in the titration graphic, and two you have found yourself. The three points are

  1. The start at $V_\text{titr}=0$
  2. Halfway to the equivalence point at $V_\text{titr}=10~\mathrm{mL}$
  3. The equivalence point at $V_\text{titr} = 20~\mathrm{mL}$

Why is number 2 important? Because that is where we can read out the $\text{p}K_\mathrm{a}$ value of the acid, because at that point $[\ce{HA}] = [\ce{A-}]$.

Now, armed with this information, here is how to proceed:

  1. Find out total amount of acid
  2. Find out total concentration of acid
  3. Calculate the initial volume from the data gathered in steps 1 and 2

1 Total amount of acid

The total amount of acid can be gleaned from information point 3, the equivalence point.

$$ n_\text{eq} = V_\text{titr,eq} \cdot c_\text{titr} = 20~\mathrm{mL} \cdot 0.150~\mathrm{mol\,L^{-1}} = 0.003~\mathrm{mol} $$

2 Total concentration of acid

The total concentration can be calculated from information points 1 and 2. We know the initial concentrations (denoted by subscript $_0$) because of the pH value at the beginning: $$ [\ce{A-}]_0 = [\ce{H+}]_0 = 10^{-3}$$

Using the equilibrium constant that we got at point two, we can calculate the initial concentration of the associated (un-dissociated) acid species: $$ [\ce{HA}]_0 = \frac{[\ce{H+}][\ce{A-}]}{K_\mathrm{a}} = \frac{10^{-6}}{10^{-5}} = 10^{-1} $$

The total initial concentration of all acid is then trivial: $$ [\ce{A}]_\text{tot,0} = [\ce{HA}]_0 + [\ce{A-}]_0 = 0.101 $$

Note that we here actually calculated the activities and not concentrations. Since we assume an activity coefficient of $\gamma = 1$ because we only consider dissolved species, the concentrations are simply the activity multiplied with the standard concentration: $$ c(\ce{A})_\text{tot,0} = \gamma \cdot [\ce{A}]_\text{tot,0} \cdot 1~\mathrm{mol\,L^{-1}} = 0.101~\mathrm{mol\,L^{-1}} $$

3 Calculation of the initial volume

The initial volume is now calculated following the simple concentration formula: $$ c_0 = \frac{n_0}{V_0} \Leftrightarrow V_0 = \frac{n_0}{c_0}$$

Since the amount of acid is constant ($n = n_0\; \mathrm{const}$), we can easily fill in the values: $$ V_0 = \frac{0.003~\mathrm{mol}}{0.101~\mathrm{mol\, L^{-1}}} \approx 0.0297~\mathrm{L} = 29.7~\mathrm{mL} $$

  • $\begingroup$ Extremely comprehensive answer, thanks. One question: what is the activities and how is it different from concentrations ? $\endgroup$
    – krismath
    Oct 9, 2014 at 14:39
  • $\begingroup$ Activity is, for the purposes here, just a fancy word for concentration. For more in-depth information, check out this link. $\endgroup$
    – tschoppi
    Oct 9, 2014 at 16:14

Here is a different approach:

From the titration curve we estimate the following:

  1. The solution of the acid has about the pH = 3 before the titration is initiated.
  2. The half titration point is at about pH = 5, i.e. equal to the pKa of the acid.
  3. At the equivalence point, 20.00 ml of 0.150 M $\ce{NaOH}$ has been added.

We look at the protolysis reaction of the acid: $$\ce{HA + H2O = A- + H3O+}$$

Before equilibrium:

  • $\mathrm{[\ce{HA}] = x}$
  • $\mathrm{[\ce{A-}] = 0}$
  • $\mathrm{[\ce{H3O+}] = 0}$

After equilibrium:

  • $\mathrm{[\ce{HA}] = x - 10^{-3}}$
  • $\mathrm{[\ce{A-}] = 10^{-3}}$
  • $\mathrm{[\ce{H3O+}] = 10^{-3}}$
  • $\mathrm{Ka = [\ce{A-}]\times[\ce{H3O+}]/[\ce{HA}] = 10^{-5}}$

Solving the equation for $x$ gives:
$\mathrm{x = \pu{0.101 M} = [\ce{HA}]}$

To convert all $\ce{HA}$ to its corresponding base $\ce{A-}$ required:

At the equivalence point $[\ce{A-}] = 20.00\cdot0.150/(20.00 + x)$, where $x$ is the volume of the acid being titrated. At the equivalence point:

$0.101 = 20.00\times0.150/(20.00 + x) \implies x = 9.7 \approx 10 ml$

Answer: The original volume of the acidic solution was 10 ml.


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