Dear Ernest Troyss Pilapil:

The salt does not possess acidic properties. Acid has acidic property.  The greater the acid dissociation constant, the greater its acidic properties. Sodium acetate and hydrochloric acid to react

H+ + CH3COO- = CH3COOH

Sodium and chlorine ions "floating" in the solution.

Best regards.

Dear Ernest

Sodium acetatein general used as buffer for adjusting the PH and as Prof.Yuri Mirgorod mentiod The greater the acid dissociation constant, the greater its acidic properties. HCl react with Sodium acetate to generate acetic acid and NaCl.

Please, find in attach, a letter on: pH of Sodium Acetate Solutions

Best widhes

Thank you for your responses Prof. Mirgorod and Prof. El-Emary.

Yes, sodium acetate can be acidified as sodium acetate has basic character. The reaction of sodium acetate with HCl  will provide acetic acid and sodium chloride.

On predicting the pH of pure sodium acetate aq. sol.; see: https://www.researchgate.net/post/Can_we_keep_the_pH_of_3M_Sodium_Acetate_around_7_instead_45-5_while_using_this_in_DNA_extraction_from_whole_blood

About predicting the pH for sodium acetate solutions with added hydrochloric acid:

We may want to discriminate three stages to break down the pH range, while simplifying pH prediction; as follows:

1st stage ― Sodium acetate is quantitatively converted to acetic acid (weak acid) and NaCl by HCl (strong acid): HCl + NaCH3COO → NaCl + CH3COOH. NaCl is a neutral salt; it does not significantly contribute for pH. HCl is consumed until the above reaction becomes complete. Only hereafter; excess HCl would occur free in solution (next stages). Before the endpoint only acetic acid and still unconverted sodium acetate contribute to pH.

2nd stage ― Excess HCl first occurs and hinders acetic acid dissociation: CH3COOH + H2O ⇌ CH3COO− + H3O+, as this equilibrium becomes largely displaced to the left. Sodium acetate was completely converted to acetic acid. Both hydrochloric acid and acetic acid contribute to pH, but the acetic acid contribution is hindered.

3rd stage ― Excess HCl becomes sufficiently high, so that It becomes reasonable to neglect the acetate hydrolysis contribution to pH. Acetic acid dissociation is largely inhibited. The pH becomes mainly contributed by hydrochloric acid .

(i) pH prediction for 1st stage:

The pH can be predicted as that for an acetic acid ― sodium acetate buffer solution, even if the solution also contains neutral salt (NaCl).The Henderson-Hasselbalch equation can, in principle, be applicable:

[H3O+] ≈ Ka·CHAc/CNaAc

pH ≈ pKa + log10(CNaAc/CHAc)

For acetic acid (denoted HAc): Ka = [Ac- ]·[H3O+]/[HAc] = 1.75·10-5 M (pKa = 4.76).

cf. my posts at: https://www.researchgate.net/post/PH-calculation-of-a-mixture-of-formic-acid-NaOH-and-water

(ii) pH prediction for 2nd stage:

For acetic acid (here denoted HAc): [HAc] = [H3O+][Ac-]/Ka ≠ 0 M.

Formal concentrations (M units) are denoted by CHAc and CNaAc, for acetic acid and the sodium acetate salt; and by CHCl and CNaCl, for HCl (excess) and NaCl. They refer to the mixed solution after the conversion: HCl + NaCH3COO → NaCl + CH3COOH.

Acetate molar balance writes (CNaAc = 0M):

CHAc = [HAc] + [Ac-] = [H3O+][Ac-]/Ka + [Ac-] = {1 + [H3O+]/Ka}·[Ac-]; from which (*): [Ac-] = CHAc·Ka/{Ka + [H3O+]}

Sodium molar balance writes (CNaAc = 0M) (*): CNaCl = [Na+]

Chloride molar balance writes (*): CHCl + CNaCl = [Cl-]

Charge balance writes: [Na+] + [H3O+] = [OH-] + [Cl-] + [Ac-]

If we can accept that pH is sufficiently acidic (pH < 6; [H3O+] > 100[OH-]) for the contribution of the water autodissociation equilibrium to be neglected:

[Na+] + [H3O+] = [Cl-] + [Ac-]

After substitution (*):

[H3O+] = CHCl + CHAc·Ka/{Ka + [H3O+]}

or:

[H3O+]2 + (Ka - CHCl)·[H3O+] = Ka(CHCl + CHAc)

what solves as:

[H3O+] = - ½(Ka - CHCl) + √{Ka(CHCl + CHAc) + ¼(Ka - CHCl)2}

(iii) pH prediction for 3rd stage:

To locate the end of the previously mentioned 2nd stage (onset of 3rd) ― based on last equation but not including HCl contribution ― we may accept that:

- ½Ka + √{CHAc·Ka + ¼(Ka)2} ≈ [H3O+]/100;

or with CHAc·Ka > 100·¼(Ka)2 and √{CHAc·Ka + ¼(Ka)2} > 100·½Ka: [H3O+] ≈ 100√{CHAc·Ka}

Hereafter, as more HCl is added, the pH can be (approximately) predicted just as:

[H3O+] ≈ CHCl; pH ≈ pCHCl = - log10(CHCl/M)

Note that the equation previously obtained for 2nd stage still holds. With sufficient added HCl; so that CHCl > 100·Ka, ¼(CHCl)2 > 100·Ka(CHCl + CHAc), and √{Ka(CHCl + CHAc) + ¼(CHCl)2} > 100½·(Ka - CHCl); it also simplifies to [H3O+] ≈ CHCl.