# Question 4b2ef

Jul 11, 2017

Dilute 24.63 ml of 5.25% NaOCl to 500 ml total volume => pH = 10 for a 0.259% NaOCl(aq) solution.

#### Explanation:

a. Determine concentration NaOCl that will give pH = 10.

$p H = 10 \implies p O H = 4 \implies \left[O {H}^{-}\right] = 1 \times {10}^{- 4}$
${K}_{a} \left(H O C l\right) \circ {25}^{o} C = 3.5 \times {10}^{-} 8$

$N a O C l \implies N {a}^{+} + O C {l}^{-}$

$\textcolor{w h i t e}{m m m m m m m m m m} O C {l}^{-} + H O H r i g h t \le f t h a r p \infty n s H O C l + O {H}^{-}$
"C_(eq)/(mol·L"^"-1"):color(white)(m)(?)color(white)(mmmmmmll)10^(-4)color(white)(mm)10^(-4)

${K}_{b} = \frac{{K}_{w}}{{K}_{a}} = \frac{\left[H O C l\right] \left[O {H}^{-}\right]}{\left[O C {l}^{-}\right]} = \frac{{\left({10}^{- 4}\right)}^{2}}{\left[O C {l}^{-}\right]} = \frac{1.0 \times {10}^{-} 14}{3.5 \times {10}^{-} 8}$

Solve for $\left[O C {l}^{-}\right] = \frac{\left(3.5 \times {10}^{-} 8\right) \left({10}^{-} 8\right)}{1.0 \times {10}^{-} 14} M = 0.035 M \text{ in" OCl^- = 0.035M "in} N a O C l$

=> 0.035M NaOCl = (0.035mol)/(Liter) => %NaOCl. "for" . pH = 10 => (0.035mol(74(g/(mol))))/(1000g)xx100% = 0.259% NaOCl

Dilution Factor = (5.25%)/(0.259%) = 20.3xx

Volume of NaOCl(5.25%) to be diluted = $\frac{500 m l}{20.3} = 24.26 m l$

Dilute 24.26 ml of 5.25% NaOCl up to but not to exceed 500 ml total volume => pH = 10 for a 0.259% NaOCl(aq) solution.

Verification:
0.259% NaOCl = 0.259% OCl^- = (0.259g)/(100g) = (((0.259g)/(74(g/(mol)))))/(0.100"Liter") = 0.035M OCl^-

$\textcolor{w h i t e}{m m m m m m m m m m} O C {l}^{-} + H O H r i g h t \le f t h a r p \infty n s H O C l + O {H}^{-}$
"C_(eq)#$\left(\frac{\text{mol}}{L}\right) : \textcolor{w h i t e}{m m m} \left(0.035 M\right) \textcolor{w h i t e}{m m m m m l l} \left(X\right) \textcolor{w h i t e}{m m} \left(X\right)$

${K}_{b} = \frac{{K}_{w}}{{K}_{a}} = \frac{\left[H O C l\right] \left[O {H}^{-}\right]}{\left[O C {l}^{-}\right]} = \left(\frac{{X}^{2}}{0.035 M} = \frac{1.0 \times {10}^{-} 14}{3.5 \times {10}^{-} 8}\right)$

$X = \left[O {H}^{-}\right] = \sqrt{\left(1.0 \times {10}^{-} 14\right) \frac{0.035}{3.5 \times {10}^{-} 8}} M = {10}^{-} 4 M$

$p O H = - \log \left[O {H}^{-}\right]$ = $- \log \left({10}^{-} 4\right) = 4$

=> $p H = 14 - p O H = 14 - 4 = 10$