a. Determine concentration NaOCl that will give pH = 10.
#pH = 10 => pOH = 4 => [OH^-] = 1xx10^(-4)#
#K_a(HOCl)@25^oC = 3.5xx10^-8#
#NaOCl => Na^+ + OCl^-#
#color(white)(mmmmmmmmmm)OCl^- + HOH rightleftharpoons HOCl + OH^-#
#"C_(eq)/(mol·L"^"-1"):color(white)(m)(?)color(white)(mmmmmmll)10^(-4)color(white)(mm)10^(-4)#
#K_b = (K_w)/(K_a) = ([HOCl][OH^-])/([OCl^-]) = ((10^(-4))^2)/([OCl^-]) = (1.0xx10^-14)/(3.5xx10^-8)#
Solve for #[OCl^-] = ((3.5xx10^-8)(10^-8))/(1.0xx10^-14)M = 0.035M" in" OCl^- = 0.035M "in" NaOCl#
=> #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 = #(500 ml)/20.3 = 24.26 ml#
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^-#
#color(white)(mmmmmmmmmm)OCl^- + HOH rightleftharpoons HOCl + OH^-#
#"C_(eq)##("mol"/L):color(white)(mmm)(0.035M)color(white)(mmmmmll)(X)color(white)(mm)(X)#
#K_b = (K_w)/(K_a) = ([HOCl][OH^-])/([OCl^-]) = ((X^2)/(0.035M) = (1.0xx10^-14)/(3.5xx10^-8))#
#X = [OH^-] = sqrt((1.0xx10^-14)(0.035)/(3.5xx10^-8))M = 10^-4M#
#pOH = -log[OH^-]# = #-log(10^-4) = 4 #
=> #pH = 14 - pOH = 14 - 4 = 10#
