Question

Question #77396

Answer 1

`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")`