calculate the percent ionization of a weak acid with a concentration of 1.95 m and a ph of…

Explanation:

Step1: Find the hydrogen - ion concentration

The formula for pH is $pH = -\log[H^+]$. Rearranging for $[H^+]$, we get $[H^+]=10^{-pH}$. Given $pH = 1.05$, then $[H^+]=10^{- 1.05}\ M$. [ [H^+]=10^{-1.05}\approx0.0891\ M ]

Step2: Calculate the percent - ionization

The formula for percent - ionization of a weak acid is $\text{Percent ionization}=\frac{[H^+]}{[HA]_0}\times100%$, where $[H^+]$ is the hydrogen - ion concentration at equilibrium and $[HA]_0$ is the initial concentration of the weak acid. Given $[HA]_0 = 1.95\ M$ and $[H^+]=0.0891\ M$. [ \text{Percent ionization}=\frac{0.0891}{1.95}\times100%=\frac{8.91}{1.95}% \approx 4.6% ]

Answer:

4.6%