Question

Question #18150

Answer 1

`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`

Answer 2

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`