The Foundations of Chemistry

(Marcin) #1
Plan
We first write the balanced equation for the reaction between NaOH and KHP. We then calcu-
late the number of moles of NaOH in 20.00 mL of solution from the amount of KHP that
reacts with it. Then we can calculate the molarity of the NaOH solution.

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Solution

NaOHKHP 88nNaKPH 2 O
1 mol 1 mol 1 mol 1 mol

We see that NaOH and KHP react in a 11 mole ratio. One mole of KHP is 204.2 g.

_?_ mol NaOH0.3641 g KHP0.001783 mol NaOH

Then we calculate the molarity of the NaOH solution.

0.08915 MNaOH

You should now work Exercise 30.

Impure samples of acids can be titrated with standard solutions of bases. The results
can be used to determine percentage purity of the samples.

EXAMPLE 11-9 Determination of Percent Acid
Oxalic acid, (COOH) 2 , is used to remove rust stains and some ink stains from fabrics. A
0.1743-gram sample of impureoxalic acid required 39.82 mL of 0.08915 MNaOH solution
for complete neutralization. No acidic impurities were present. Calculate the percentage purity
of the (COOH) 2.
Plan
We write the balanced equation for the reaction and calculate the number of moles of NaOH
in the standard solution. Then we calculate the mass of (COOH) 2 in the sample, which gives
us the information we need to calculate percentage purity.
Solution
The equation for the complete neutralization of (COOH) 2 with NaOH is

2NaOH(COOH) 2 88nNa 2 (COO) 2 2H 2 O
2 mol 1 mol 1 mol 2 mol

Twomoles of NaOH neutralizes completely one mole of (COOH) 2. The number of moles of
NaOH that react is the volume times the molarity of the solution.

_?_ mol NaOH0.03982 L0.003550 mol NaOH

Now we calculate the mass of (COOH) 2 that reacts with 0.003550 mol NaOH.

0.08915 mol NaOH

L

0.001783 mol NaOH

0.02000 L

_?_ mol NaOH

L

1 mol NaOH

1 mol KHP

1 mol KHP

204.2 g KHP
Very pure KHP is readily available.


410 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations


Each molecule of (COOH) 2 contains
two acidic H’s.


C

O

H O C O H

O

1 mol  90.04 g
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