Question #18150
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I will assume that this is at 298K.
#500mL=0.5L=0.5dm^3#
#n(NaOH)=0.0125/40=3.125*10^(-4)# #mol#
#[NaOH]=(3.125*10^(-4))/0.5=6.25*10^(-4)# #mol# #dm^(-3)#
Since NaOH is a strong base we will assume it fully dissociates.
#[OH^-]=6.25*10^(-4)# #mol# #dm^(-3)#
#[H^+]=(K_w)/([OH^-])=(1*10^(-14))/(6.25*10^(-4))=1.6*10^(-11)# #mol# #dm^(-3)#
#pH=-log([H^+])=-log(1.6*10^(-11))~~10.8#
First find the molarity of 0.0125g of 500mL NaOH solution
#M=color(red)("Number of MOLES of solute")/color(red)("Litre")#
#M=(0.0125/40)/(0.5L) "moles" rArr "0.000625 moles per Litre"#
#color(red)rArr6.25xx10^-4# #"mol/L"#
The concentration of #OH^-# ions #color(red)rArr6.25xx10^-4# #"mol/L"#
#p(OH)=-log{OH^(-) " concentration"}#
#p(OH)=-log{6.25xx10^-4 }#
#p(OH)=4-log{6.25}#
#p(OH)=4-0.7958#
#p(OH)=3.2042rArr3.2#
Since
#p(OH)+p(H)=14#
you have
#pH = 14 - p(OH) rArr 14 - 3.2 = 10.8#
