I had this problem too on my Lenovo Thinkpad due to ppa graphics dirvers. I solved it by installing ppa-purge and then removing the https://launchpad.net/~oibaf/+archive/graphics-drivers ppa:


sudo apt-get install ppa-purge

sudo ppa-purge ppa:oibaf/graphics-drivers


Upgrade to 12.04 Failed due to held back packages
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 millilitres 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: 
$$ \ce{HA_{(aq)} + H2O_{(l)} &lt;=&gt; A_{(aq)}- + H3O+_{(aq)} } $$

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 litre of the solution, there will be exactly 0.150 mole 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 $\ce{OH^-}$

