# Finding the pH value of HF [closed]

Find the pH value of HF.

$pH = -\log[H+]$ for strong acids

$pOH = -\log[OH-]$ for strong bases

$pH = 1/2(pKa - \log C)$ for weak acid

$pOH = 1/2(pKb - \log C)$ for weak base

HF is weak acid so $Ka = 6.6 \times 10^{-4}$

$pH = 1/2(-\log(6.6 \times 10^{-4}) - \log(0.04))$

$\boxed {pH = 2.2892}$

Is my approach correct?

## closed as unclear what you're asking by Mithoron, MaxW, Jon Custer, Todd Minehardt, airhuffJun 17 '18 at 2:05

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• This is complete nonsense. You need concentration to calculate pH. – Mithoron Jun 16 '18 at 16:15
• Have a look at my answer to this question as it outlines the full method and its approximations chemistry.stackexchange.com/questions/60068/… – porphyrin Jun 17 '18 at 8:49

Well, yes but barely. Assuming you forgot to say that $C = 0.04M$ is the molarity of the HF solution, then you have applied your equations correctly.
A concern here is that your shortcut equations rely on what's often called the "small x approximation," which says the amount of HF that actually dissociates is so small we can neglect the difference between the initial and equilibrium molarity of HF. Is that actually true? We would need to do the full calculation to see, starting from the Ka equation that describes the equilibrium composition of the solution: $$K_a = \frac{[{\rm H}_3{\rm O}^{+}][{\rm F}^{-}]}{[{\rm HF}]}$$ If we let $x$ stand for the equilibrium molarity of ${\rm H}_3{\rm O}^{+}$ and $C$ for the initial molarity of HF, then this equation turns into $$x^2 = K_a (C - x)$$ which can be solved with the quadratic equation. Using $C = 0.04$ we find $x = 4.819\times10^{-3}$ which gives pH = 2.317... You'll notice that's different from your value, and that's because in fact about 12% of the HF dissociates, which isn't that small.