Question:
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.
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 $\ce{OH^-}$