a. Determine concentration NaOCl that will give pH = 10.
pH = 10 => pOH = 4 => [OH^-] = 1xx10^(-4)pH=10⇒pOH=4⇒[OH−]=1×10−4
K_a(HOCl)@25^oC = 3.5xx10^-8Ka(HOCl)∘25oC=3.5×10−8
NaOCl => Na^+ + OCl^-NaOCl⇒Na++OCl−
color(white)(mmmmmmmmmm)OCl^- + HOH rightleftharpoons HOCl + OH^-mmmmmmmmmmOCl−+HOH⇌HOCl+OH−
"C_(eq)/(mol·L"^"-1"):color(white)(m)(?)color(white)(mmmmmmll)10^(-4)color(white)(mm)10^(-4)C_(eq)/(mol⋅L-1):m(?)mmmmmmll10−4mm10−4
K_b = (K_w)/(K_a) = ([HOCl][OH^-])/([OCl^-]) = ((10^(-4))^2)/([OCl^-]) = (1.0xx10^-14)/(3.5xx10^-8)Kb=KwKa=[HOCl][OH−][OCl−]=(10−4)2[OCl−]=1.0×10−143.5×10−8
Solve for [OCl^-] = ((3.5xx10^-8)(10^-8))/(1.0xx10^-14)M = 0.035M" in" OCl^- = 0.035M "in" NaOCl[OCl−]=(3.5×10−8)(10−8)1.0×10−14M=0.035M inOCl−=0.035MinNaOCl
=> 0.035M NaOCl = (0.035mol)/(Liter)
=> %NaOCl. "for" . pH = 100.035MNaOCl=0.035molLiter⇒%NaOCl.for.pH=10 => (0.035mol(74(g/(mol))))/(1000g)xx100%0.035mol(74(gmol))1000g×100% = 0.259% NaOCl0.259%NaOCl
Dilution Factor = (5.25%)/(0.259%) = 20.3xx5.25%0.259%=20.3×
Volume of NaOCl(5.25%) to be diluted = (500 ml)/20.3 = 24.26 ml500ml20.3=24.26ml
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^-0.259%NaOCl=0.259%OCl−=0.259g100g=(0.259g74(gmol))0.100Liter=0.035MOCl−
color(white)(mmmmmmmmmm)OCl^- + HOH rightleftharpoons HOCl + OH^-mmmmmmmmmmOCl−+HOH⇌HOCl+OH−
"C_(eq)Ceq("mol"/L):color(white)(mmm)(0.035M)color(white)(mmmmmll)(X)color(white)(mm)(X)(molL):mmm(0.035M)mmmmmll(X)mm(X)
K_b = (K_w)/(K_a) = ([HOCl][OH^-])/([OCl^-]) = ((X^2)/(0.035M) = (1.0xx10^-14)/(3.5xx10^-8))Kb=KwKa=[HOCl][OH−][OCl−]=(X20.035M=1.0×10−143.5×10−8)
X = [OH^-] = sqrt((1.0xx10^-14)(0.035)/(3.5xx10^-8))M = 10^-4MX=[OH−]=√(1.0×10−14)0.0353.5×10−8M=10−4M
pOH = -log[OH^-]pOH=−log[OH−] = -log(10^-4) = 4 −log(10−4)=4
=> pH = 14 - pOH = 14 - 4 = 10pH=14−pOH=14−4=10
