Question #18150

2 Answers
1s2s2p
Dec 5, 2017

pH≈10.8

Explanation:

I will assume that this is at 298K.

500mL=0.5L=0.5dm3

n(NaOH)=0.012540=3.125⋅10−4 mol

[NaOH]=3.125⋅10−40.5=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+]=Kw[OH−]=1⋅10−146.25⋅10−4=1.6⋅10−11 mol dm−3

pH=−log([H+])=−log(1.6⋅10−11)≈10.8

Milan S# · Stefan V.
Dec 5, 2017

pH=10.8

Explanation:

First find the molarity of 0.0125g of 500mL NaOH solution

M=Number of MOLES of soluteLitre

M=0.0125400.5Lmoles⇒0.000625 moles per Litre

⇒6.25×10−4 mol/L

The concentration of OH− ions ⇒6.25×10−4 mol/L

p(OH)=−log{OH− concentration}

p(OH)=−log{6.25×10−4}

p(OH)=4−log{6.25}

p(OH)=4−0.7958

p(OH)=3.2042⇒3.2

Since

p(OH)+p(H)=14

you have

pH=14−p(OH)⇒14−3.2=10.8

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