calculate the percent ionization of a weak acid with a concentration of 1.75 m and a ph of…

Explanation:

Step1: Calculate [H⁺] from pH

The formula for pH is $pH = -\log[H^{+}]$. Rearranging for $[H^{+}]$ gives $[H^{+}]=10^{-pH}$. Substituting $pH = 1.35$, we have $[H^{+}]=10^{- 1.35}\ M$. $[H^{+}]=4.47\times10^{-2}\ M$

Step2: Calculate percent - ionization

The formula for percent - ionization of a weak acid is $\text{Percent ionization}=\frac{[H^{+}]}{[HA]_0}\times100%$, where $[HA]_0$ is the initial concentration of the weak acid. Given $[HA]_0 = 1.75\ M$ and $[H^{+}]=4.47\times10^{-2}\ M$. $\text{Percent ionization}=\frac{4.47\times 10^{-2}\ M}{1.75\ M}\times100%$ $\text{Percent ionization}=\frac{4.47\times10^{-2}}{1.75}\times100%=2.55% \approx 2.6%$

Answer:

$2.6%$ (corresponding to the first option in the multiple - choice list)