Part 1: For region 1 in the graph above (the weak acid before any strong base is

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

Explanation / 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.

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