## anonymous one year ago A solution containing HCl would likely have

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1. abb0t

a pH < 7

2. JFraser

A burning taste

3. Ciarán95

As @abbot implied, it would likely have a relatively low pH due to the high concentration of H+ ions present in solution. Hydrochloric acid, like all acids, will dissociate in solution to produce H+ ions and a remaining conjugate base with an overall negative charge. In this case, the conjugate base is the remaining Cl- ion, once the H+ is removed. We can represent this dissociation as part of an equilibrium system: $HCl _{(aq)}\rightarrow \leftarrow H _{(aq)}^{+} + Cl _{(aq)}^{-}$ The stronger an acid, the more likely it is to dissociate and lose a proton (i.e. from the products on the right-hand side in the forward reaction). This leads to the formation of the ions being more favorable in solution rather than HCl remaining 'together' as such. Furthermore, the stronger an acid is, the subsequently weaker its conjugate base will be (in it's basicity/ its ability to accept a H+ ion back and form HCl once more). The Cl- ion is a chlorine atom with a full octet (outer shell of electrons), making it relatively very stable and unlikely to want to try and share any of these electrons by bonding to a proton - they are kind of happy floating about on their own in solution! Hope that helps! :)

4. Ciarán95

By the way @rommerro ..... $$\large Welcome~to~OpenStudy!~:)$$

5. Photon336

even so as @Ciarán95 pointed out another thing to support what he said the acid's ka would be significantly greater than one like <<<<<1 because virtually all the acid would be dissociated$K _{a} = \frac{ [H]^{+}[A ^{-} ] }{ [HA] }$. the strength of the acid is inversely related to the conjugate base. stronger acid = weaker conjugate base. Makes sense because the conjugate base would virtually have no affinity for that proton.

6. abb0t

How would the pKa be greater than 1 a strong acid such as HCl? Please, do explain that.

7. Photon336

Ka for HCL was like 10^6 so pKa would be less than one not greater than one Sorry about that!