Question #9e199

Apr 27, 2017

How to do the problem appears below...

Explanation:

In each case, the relation you want to use is:

${M}_{a} {V}_{a} = {M}_{b} {V}_{b}$

which applies at the endpoint of a titration.

However, you must realize that this equation really says:

"moles of acid = moles of base" at the equivalence point.

This is true for items (a) and (b) since the acids are monoprotic - they have a single hydrogen ion to donate.

In (c), you have a triprotic acid (three protons transferred) and in (d) it is diprotic (two protons).

This means the amount of base needed must be triple (in c) and double (in d) the amount of acid used.

You can modify the formula by multiplying the left side (${M}_{a} {V}_{a}$) by three and by two in (c) and (d) respectively.

For example:

c) $3 \times \left(0.134\right) \left(10.6\right) = \left(1.40\right) {V}_{b}$

${V}_{b} = \frac{3 \times \left(0.134\right) \left(10.6\right)}{1.40} = 3.04 m L$

Notice that this is a rare instance where it is not necessary to convert volume to litres.

For (a) and (b) omit the "3", and in (d) use a 2 instead.