## Chemistry: The Central Science (13th Edition)

$[OH^-]= 6.325 \times 10^{- 4}M$ $pH = 10.80$
- 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: $pOH = -log[OH^-]$ $pOH = -log( 6.325 \times 10^{- 4})$ $pOH = 3.20$ $pH + pOH = 14$ $pH + 3.199 = 14$ $pH = 10.80$ 