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Untitled - Kelly Walsh High School

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72 CHEMISTRY FOR THE UTTERLY CONFUSED<br />

All possible titration problems will simply be adding steps before and/or after<br />

one of these two mole ratio terms. The problem will begin with the substance<br />

with the most information, and proceed through one of these mole ratios to the<br />

substance with less information given.<br />

Let’s begin with this titration question: How many moles of calcium hydroxide<br />

are necessary to titrate 0.250 mol of acetic acid? We have information about the<br />

acetic acid, and we are seeking information about the calcium hydroxide. We<br />

will begin the problem with the acetic acid since we know more about it, and we<br />

will end the problem with the calcium hydroxide since we do not know anything<br />

about this compound. One way to remember what you need to do is to copy the<br />

given information and the question to the balanced chemical equation, as<br />

shown here:<br />

2 HC2H3O2(aq) Ca(OH) 2(aq) l Ca(C2H3O2) 2(aq) 2 H2O(l) 0.250 mol ? mol<br />

When written this way, you begin with the number and work towards the question<br />

mark. We begin with the acetic acid and use the mole ratio with acetic acid<br />

in the denominator:<br />

(0.250 mol HC2H3O2 ) a 1 mol Ca(OH) 2<br />

b 0.125 mol Ca(OH) 2<br />

2 mol HC2H3O2 This calculation leaves us with the appropriate number of significant figures.<br />

Note that the units of HC 2H 3O 2 have cancelled, leaving the desired units of<br />

mol Ca(OH) 2.<br />

How would the problem be different if the titration question asked is: How<br />

many moles of calcium hydroxide are necessary to titrate 0.0500 L of a 0.100 M<br />

acetic acid solution? We can begin by adding our information to the balanced<br />

chemical equation:<br />

2 HC 2H 3O 2(aq) Ca(OH) 2(aq) l Ca(C 2H 3O 2) 2(aq) 2 H 2O(l)<br />

0.0500 L ? mol<br />

0.100 M<br />

As in the preceding example, we will begin with the acetic acid and go through<br />

the same mole ratio to get to the calcium hydroxide. The difference in this problem<br />

is that we are beginning with liters and molarity instead of moles. We will<br />

need to add a step to our calculation to change liters and molarity to moles. Part

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