Question #18150

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Answer 1 · 2017-12-05T11:40:59

$pH~~10.8$

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 · 2017-12-05T12:05:38

pH $= 10.8$

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

$pH = 14 - p(OH) rArr 14 - 3.2 = 10.8$