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, airhuff Jun 17 '18 at 2:05

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ This is complete nonsense. You need concentration to calculate pH. $\endgroup$ – Mithoron Jun 16 '18 at 16:15
  • $\begingroup$ Have a look at my answer to this question as it outlines the full method and its approximations chemistry.stackexchange.com/questions/60068/… $\endgroup$ – 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.

However, a point we haven't addressed yet is that you've given your answer to far more precision that the data justify. You have only 2 significant digits in your pKa and only 1 in your value for C. That means your answer should have only one significant digit, and you should give it as pH = 2. In that case, taking the precision of your data into account, your answer is correct. In fact, your answer is correct even to two significant digits (pH = 2.3). But that's kind of just lucky. You might want to study up on when you can use those equations and when you can't.


Not the answer you're looking for? Browse other questions tagged or ask your own question.