Question #18150

2 Answers
Dec 5, 2017

Answer:

#pH~~10.8#

Explanation:

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#

Dec 5, 2017

Answer:

pH#= 10.8#

Explanation:

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#