- Now, the equilibrium constant,
becomes![=\frac{x\cdot x}{0.256-x}](http://latex.codecogs.com/gif.latex?=\frac{x\cdot&space;x}{0.256-x})
- Plugging in values,
![1.34\times 10^{-5}=\frac{x\cdot x}{0.256-x}](http://latex.codecogs.com/gif.latex?1.34\times&space;10^{-5}=\frac{x\cdot&space;x}{0.256-x})
- Remember that in weak acids, x is very very small. How do we know that it is a weak acid? By the fact that it has an equilibrium constant. Strong acids are “single arrow” reactions, that means there is no equilibrium or equilibrium constant associated.
- So what does this mean? It means that 0.265 – x is approximtely equal to 0.265!! That is incredibly cool, because we just converted a messy quadratic into a LINEAR function.
- Let us now solve
, or![x=1.88\times 10^{-3}](http://latex.codecogs.com/gif.latex?x=1.88\times&space;10^{-3})
- Time to put on units.
M. CONTEXT: we are looking for H+.
- And… we all know that pH = -log [H+]. Plugging in, doing the math,
![pH=-log (1.88\times 10^{-3}) = 2.725](http://latex.codecogs.com/gif.latex?\inline&space;pH=-log&space;(1.88\times&space;10^{-3})&space;=&space;2.725)
I know that many of you haven’t gotten to buffers yet, so I decided to truncate this problem with sections (a) and (b). This is the time to post questions, even if they are just about significant digits! Good luck, and ask away.