Answer
$pH = 10.30$
Work Step by Step
- Identify the electrolytes:
$Na^+$: Insignificant
$ClO^-$: Weak base
$K^+$: Insignificant
$I^-$: Insignificant
- Therefore, we just need to consider the $ClO^-$ to calculate the pH.
1. Since $ClO^-$ is the conjugate base of $HClO$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 2.9\times 10^{- 8} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 2.9\times 10^{- 8}}$
$K_b = 3.448\times 10^{- 7}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [HClO] = x$
-$[ClO^-] = [ClO^-]_{initial} - x = 0.115 - x$
For approximation, we consider: $[ClO^-] = 0.115M$
3. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][HClO]}{ [ClO^-]}$
$Kb = 3.448 \times 10^{- 7}= \frac{x * x}{ 0.115}$
$Kb = 3.448 \times 10^{- 7}= \frac{x^2}{ 0.115}$
$ 3.966 \times 10^{- 8} = x^2$
$x = 1.991 \times 10^{- 4}$
Percent ionization: $\frac{ 1.991 \times 10^{- 4}}{ 0.115} \times 100\% = 0.1732\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [HClO] = x = 1.991 \times 10^{- 4}M $
$[ClO^-] \approx 0.115M$
4. Calculate the pH.
$pOH = -log[OH^-]$
$pOH = -log( 1.991 \times 10^{- 4})$
$pOH = 3.701$
$pH + pOH = 14$
$pH + 3.701 = 14$
$pH = 10.299$
