Question #77396

1 Answer
Jun 3, 2017

Answer:

#sf((a))#

4.4

#sf((b))#

4.9

#sf((c))#

#sf(7.9xx10^(-10)color(white)(x)"mol/l")#

Explanation:

#sf((a))#

#sf(pH=-log[H^+])#

At this low concentration I thought that you may need to account for the #sf(H^+)# ions which arise from the dissociation of water. I ran it through a calculation and found that this has negligible effect.

500 ml = 0.5 L

#sf([H^+]=c/v = (2.0xx10^(-5))/(0.5)=4xx10^(-5)color(white)(x)"mol/l")#

#:.##sf(pH=-log(4xx10^(-5))=4.4)#

#sf((b))#

#sf([H^+]=(2.0xx10^(-5))/(1.5)=1.3xx10^(-5)color(white)(x)"mol/l")#

#:.##sf(pH=-log(1.3xx10^(-5))=4.9)#

#sf((c))#

#sf(pH+pOH=14)#

#:.##sf(pOH=14-pH=14-4.9=9.1)#

#sf([OH^-]=10^(-9.1)=7.9xx10^(-10)color(white)(x)"mol/l")#