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#1
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what would be the ph of s solution made by dissolving .1 mole of acetic acid and .1 mole of sodium acetate in a 1 liter of water?
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#2
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NaAc---->(Na+)+(Ac-)
1mol/L 1mol/L 1mol/L HAc<==>(H+)+(Ac-) C(before reaction) 1 0 1 C(when balance) 1-x x 1+x K[HAc]= C(H+).C(Ac-)/C(HAc) Look up the table,find K[HAc]=1.8*0.00001 0.000018=x(1+x)/(1-x) x is tiny, so we can regard 1+x & 1-x as 1 therefore C(H+)=x=0.000018mol/L pH=-lg[C(H+)]=-lg0.000018=4.74 |
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#3
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I've always found the Henderson-Hasselbalch equation a simple way to determine ph of a solution knowing the moles of weak acid and conjugate base.
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