Answer :
The % acid dissociation of a 1.20M solution of hydrofluoric acid with pH=1.55 is 4.77%.
To find the percent acid dissociation, we need to calculate the concentration of hydrogen ions [tex](\( [\text{H}^+] \))[/tex] in the solution first. Given that [tex]pH=-\(\log[\text{H}^+]\),[/tex] we can rearrange the equation to solve for [tex]\( [\text{H}^+] \) which gives \( [\text{H}^+] = 10^{-\text{pH}} \).[/tex] Substituting the given pH value of 1.55, we get [tex]\( [\text{H}^+] = 10^{-1.55} \).[/tex]Calculating this, we find [tex]\( [\text{H}^+] = 3.548 \times 10^{-2} \).[/tex]
Next, we need to find the initial concentration of [tex]HF (\( [\text{HF}]_0 \)) using the formula \( \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]_0}\right) \), where \(\text{A}^-\)[/tex] is the concentration of the conjugate base and [tex]\(\text{HA}\)[/tex] is the concentration of the acid. Given that HF dissociates as [tex]HF \rightleftharpoons \text{H}^+ + \text{F}^- \)[/tex], the initial concentration of [tex]\( [\text{HF}]_0 \)[/tex] is the same as the molarity of the solution, which is 1.20M.
Now, we can use the equation for percent dissociation: % dissociation =[tex]\( \frac{[\text{H}^+]}{[\text{HF}]_0} \times 100 \)[/tex]. Substituting the calculated values,
we get % dissociation =[tex]\( \frac{3.548 \times 10^{-2}}{1.20} \times 100 \).[/tex]Solving this, we find the percent dissociation to be approximately 4.77%.
