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From my textbook, it is stated that:

e.g. If we add a basic salt $\ce{NaF}$, it will fully dissociate to give $\ce{F-}$. Then if we add $\ce{HF}$, since it is a weak acid, the high concentration of $\ce{F-}$ already present in solution shifts the equilibrium to the left, hence fewer $\ce{H+}$ forms than expected so higher $\mathrm{pH}$.

But can I explain like this: after dissociation of $\ce{NaF}$, the formed $\ce{F-}$ reacts with water to give $\ce{OH-}$ as the following equation: $$\ce{F- + H2O <=> HF + OH-}$$. Due to the presence of $\ce{OH-}$, upon addition of $\ce{HF}$, the $\mathrm{pH}$ will be higher than expected. Are there any flaws in this explanation?

Adnan AL-Amleh
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    You picked a really bad example. For example for KCN both paragraphs would be OK. For HF thing look very differently https://chemistry.stackexchange.com/a/34829/9961 – Mithoron Jan 31 '21 at 20:52
  • I agree, with forming HF2- and rapid increasing of HF acidity for concentrated solutions. ( $\ce{3 HF <=> H2F+ + HF2- }$ ) – Poutnik Jan 31 '21 at 22:17

1 Answers1

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Both your paragraphs are valid, as all present components are balanced by 3, mutually linked equilibrium reactions, with the respective equilibrium constants:

$\ce{HF + H2O <=> F- + H3O+}$, $K_\mathrm{a}$

$\ce{F- + H2O <=> HF + OH-}$, $K_\mathrm{b}$

$\ce{2 H2O <=> H2O+ + OH-}$, $K_\mathrm{w}$

with the relation $K_\mathrm{a} \cdot K_\mathrm{b} = K_\mathrm{w}$

$\mathrm{pH}= \mathrm{p}K_\mathrm{a} + \log{\frac{\ce{[F-]}}{\ce{[HF]}}}$

Poutnik
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  • thanks. To clarify, so both can explain the effect, and are there any causal relationship between the 2 explanations since you said they are 'linked'? – Question Jan 31 '21 at 15:13
  • Rather 2 complemental sides of the same thing. Like 2 pieces of torn up paper sheet. What is the causal relation between them ? – Poutnik Jan 31 '21 at 15:42