Question

Calculate the pH of a 0.010 M solution of iodic acid (HIO₃, K_a = 0.17).

Short answer

HIO₃ solution dissociates into H⁺ and IO₃^- ions.

pH of a 0.010 M solution of monochloroacetic acid = 2

Step-by-step solution

Calculating the Amount Dissociated

Iodic acid i.e, HIO₃solutiondissociates as,

$\begin{array}{l}{{\text{HIO}}_3}_{\text{(aq)}}\rightleftarrows {{\text{\hspace{0.33em}H}}^{\text{+}}}_{\text{(aq)}}{\text{+IO}}_3{{}^{\text{-}}}_{\text{(aq)}}\\ 0.01000\\ \text{0.010-x\hspace{0.33em}\hspace{0.33em}x\hspace{1em}\hspace{1em}\hspace{1em}x}\end{array}$

$\begin{array}{l}{\text{K}}_a=\frac{{\text{x}}^2}{0.010-x}=0.17\\ {\text{x}}^2=0.0017-0.17\text{x}\\ x^2+0.17\text{x\hspace{0.33em}}-0.0017=0\Rightarrow \text{x}=0.01\text{\hspace{0.33em}M}\end{array}$

(K_a is not small, so any approximation cannot be made here.)

Calculating pH

x is the amount dissociated or the amount of [H⁺] and [IO₃^-] ions formed.

$\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]=\text{x}=0.01\\ \text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]\\ \text{pH}=-\log \left(0.01\right)\\ \text{pH}=2\end{array}$