question 3 (6 points)\nwhat is the ph of 0.45 mol/l solution of hypochlorous acid?\n(k = 3.0 x 10^{-8})\nph…

Explanation:

Step1: Write the dissociation equation

For hypochlorous acid ($HClO$), the dissociation is $HClO\rightleftharpoons H^++ClO^-$. Let $[H^+]=x$. Then at equilibrium, $[HClO]=0.45 - x$ and $[ClO^-]=x$. Since $K_a$ is very small ($3.0\times 10^{-8}$), we can assume $0.45 - x\approx0.45$.

Step2: Write the expression for $K_a$

$K_a=\frac{[H^+][ClO^-]}{[HClO]}$. Substituting the values, we have $3.0\times 10^{-8}=\frac{x\cdot x}{0.45}$.

Step3: Solve for $x$ (which is $[H^+]$)

$x^2 = 3.0\times 10^{-8}\times0.45$. So $x=\sqrt{3.0\times 10^{-8}\times0.45}=\sqrt{1.35\times 10^{-8}}\approx1.16\times 10^{-4}\ mol/L$.

Step4: Calculate the pH

$pH =-\log[H^+]$. Substituting $[H^+]=1.16\times 10^{-4}$, we get $pH=-\log(1.16\times 10^{-4})\approx 3.9$.

Answer:

pH = 3.9