Gen Chem question

[HdK]

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I got this right on bootcamp, but I did a completely different method. I want to know if my way is actually alright or if i just got lucky. this is the question and answer.

Question: What is the pH of a solution prepared by adding 20 mL of 0.02 M NaOH(aq) to 20 mL of water?

Answer:

NaOH is a strong base and fully dissociates into Na+ and OH-. To find the new concentration of NaOH after we dilute it with water, we use:

M1V1 = M2V2

(0.02)(0.02) = M2(0.04)

0.01 = M2

Notice that we use 0.04 L for the second volume, NOT 0.02 L. This is because we are adding 0.02 L of NaOH(aq) to 0.02 L of water for a total of 0.04 L.

Next, since NaOH is a strong base, we can conclude [OH-] = 0.01 M.

pOH = -log([OH-])
pOH = -log(0.01)
pOH = 2

pH + pOH = 14
pH = 14 – pOH
pH = 14 – 2 = 12


MY LOGIC:

So..I got the right answer, but i just want to know if I just got lucky. I said that since it's a strong base, it'll disassociate 100%, so I just took 0.02M as the [OH-]. So I estimated the pOH to be around 1.9 through estimating logarithms and it being closer to 2 than 1. Then just subtracted 1.9 from 14, which was closer to 12, but not exactly 12. Is this ok too..?

 

[Piepiesuperpie]

You got lucky here doing with your method. Your way does not consider the fact that it's being diluted. Here it estimated out ok because it's not a huge dilution and the pH scale increases not linearly, but rather logarithmically. Had it been diluted by 200mL of water (a larger dilution), your method would not work.