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Step-by-step calculations for determining the ph of solutions with different concentrations of ba(oh)2, hno2, and acetic acid using various methods. It covers strong and weak bases and acids, and the use of the henderson-hasselbalch equation for buffer solutions.
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CHEM1002 2008 - N- 5 November 2008
(aq)] = 0.040 M. By definition, pOH = - log 10 [OH-(aq)] so pOH = - log 10 (0.040) = 1.40. As pH + pOH = 14.00, pH = 14.00 โ 1.40 = 12. pH = 12.
initial 0.150 large 0 0 change - x negligible + x + x final 0.150 โ x large x x The equilibrium constant K a is given by: K a = ๐๐๐!^ [๐๐๐!] [๐๐๐๐]
๐๐ ๐.๐๐๐!๐ As p K a = - log 10 K a, K a = 10โ^ 3.15^ and is very small, 0.150 โ x ~ 0.150 and hence: x 2 = 0.150 ร 10
CHEM1002 2008 - N- 5 November 2008
The solution contains both a weak acid (acetic acid) and its conjugate base (the acetate ion). This is a buffer and its pH can be calculate using the Henderson- Hasselbalch equation. With [acid] = 0.080 M and [base] = 0.160 M, pH = p K a + log [๐๐๐ฌ๐] [๐๐๐ข๐] = 4.76 + log ๐.๐๐๐ ๐.๐๐๐
pH = 5.