a. Determine concentration NaOCl that will give pH = 10.
pH=10⇒pOH=4⇒[OH−]=1×10−4
Ka(HOCl)∘25oC=3.5×10−8
NaOCl⇒Na++OCl−
mmmmmmmmmmOCl−+HOH⇌HOCl+OH−
C_(eq)/(mol⋅L-1):m(?)mmmmmmll10−4mm10−4
Kb=KwKa=[HOCl][OH−][OCl−]=(10−4)2[OCl−]=1.0×10−143.5×10−8
Solve for [OCl−]=(3.5×10−8)(10−8)1.0×10−14M=0.035M inOCl−=0.035MinNaOCl
=> 0.035MNaOCl=0.035molLiter⇒%NaOCl.for.pH=10 => 0.035mol(74(gmol))1000g×100% = 0.259%NaOCl
Dilution Factor = 5.25%0.259%=20.3×
Volume of NaOCl(5.25%) to be diluted = 500ml20.3=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−=0.259g100g=(0.259g74(gmol))0.100Liter=0.035MOCl−
mmmmmmmmmmOCl−+HOH⇌HOCl+OH−
Ceq(molL):mmm(0.035M)mmmmmll(X)mm(X)
Kb=KwKa=[HOCl][OH−][OCl−]=(X20.035M=1.0×10−143.5×10−8)
X=[OH−]=√(1.0×10−14)0.0353.5×10−8M=10−4M
pOH=−log[OH−] = −log(10−4)=4
=> pH=14−pOH=14−4=10
