Part 1: For region 1 in the graph above (the weak acid before any strong base is
ID: 498378 • Letter: P
Question
Part 1:
For region 1 in the graph above (the weak acid before any strong base is added), which of the following calculations would you do?
A. Ka = x2/[HA], pH = -log[H+]
B. pH = pKa + log([A-]/[HA])
C. pH = pKa
D. pOH = -log[OH-], pH = 14-pOH
E. Kb = x2/[A-], pOH = -log[OH-], pH = 14-pH
Part 2:
For region 3 in the graph above (the weak acid with less than an equivalent amount of strong base added), which of the following calculations would you do?
A. Ka = x2/[HA], pH = -log[H+]
B. pH = pKa + log([A-]/[HA])
C. pH = pKa
D. pOH = -log[OH-], pH = 14-pOH
E. Kb = x2/[A-], pOH = -log[OH-], pH = 14-pH
Part 3:
For region 4 in the graph above (the equivalence point), which of the following calculations would you do?
A. Ka = x2/[HA], pH = -log[H+]
B. pH = pKa + log([A-]/[HA])
C. pH = pKa
D. pOH = -log[OH-], pH = 14-pOH
E. Kb = x2/[A-], pOH = -log[OH-], pH = 14-pH
14 12 10 BL equivalence point 7 & 7 half-eq point 20 10 Volume of base added (mLO 30Explanation / Answer
Part1, A. Ka=x2/[HA]. PH=-log[H=] should be used for calculation.
Part 2, B. PH=PKa+log([A-]/[HA] will be used for calculation. THis is because less than 1 equivalent strong base was added which will form salt with the weak acid and there will be excess weak acid. Thus it's a acid buffer and accordingly the formula of Henderson-Heselbach equation is used.
Part3, D.POH=-log[OH-], PH=14-PH will be used as above neutralisation point there is no more acid left, only excess NaOH is present, So for calculation this formula should be used.