J.R. S. answered • 10/28/18
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HCl + naOH ==> NaCl + H2O
moles HCl = 20.5 ml x 1L/1000ml x 0.3 mol/L = 0.00615 mol
moke NaOH = 20.0 ml x1L/1000 ml x 0.3 mol/L = 0.006 mol
mokes HCl remaining after reacty = 0.00615 - 0.006 = 0.00015
Final volume = 20.5 + 20.0 = 40.5 ml = 0.0405 L
Final [HCl] = 0.00015 mol/0.0405 L = 0.00370 M
pH = -log 0.00370 M = 2.43
J.R. S.
tutor
Maybe they want pH = 2.5 to only 2 significant figures, based on 2 sig figs in 0.30 M. Try that.
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10/28/18
Sih B.
I asked my professor the answer was 2.3. Not really sure how..
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10/28/18
J.R. S.
tutor
I had used an incorrect volume which I have corrected. The answer still does not come out 2.3. You should ask your prof to explain how that is the case. I don't think rounding errors should account for it.
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10/29/18
Sih B.
This is what I got too and my program said It was wrong10/28/18