Answer
$[OH^-]= 6.325 \times 10^{- 4}M$
$pH = 10.80$
Work Step by Step
- Identify the significant electrolytes in $NaOBr$
- $Na^+$: Insignificant acid.
- $BrO^-$: Weak base.
Therefore, to calculate the pH, we just need to consider a $0.10M$ $BrO^-$ solution.
1. Since $BrO^-$ is the conjugate base of $HBrO$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 2.5\times 10^{- 9} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 2.5\times 10^{- 9}}$
$K_b = 4.0\times 10^{- 6}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [HBrO] = x$
-$[BrO^-] = [BrO^-]_{initial} - x = 0.1 - x$
For approximation, we consider: $[BrO^-] = 0.1M$
3. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][HBrO]}{ [BrO^-]}$
$Kb = 4 \times 10^{- 6}= \frac{x * x}{ 0.1}$
$Kb = 4 \times 10^{- 6}= \frac{x^2}{ 0.1}$
$ 4 \times 10^{- 7} = x^2$
$x = 6.325 \times 10^{- 4}$
Percent ionization: $\frac{ 6.325 \times 10^{- 4}}{ 0.1} \times 100\% = 0.6325\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [HBrO] = x = 6.325 \times 10^{- 4}M $
$[BrO^-] \approx 0.1M$
4. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 6.325 \times 10^{- 4})$
$pOH = 3.20$
$pH + pOH = 14$
$pH + 3.199 = 14$
$pH = 10.80$