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High School Chemistry : Titrations

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Example Questions

Example Question #3 : Titrations And Indicators

You have a solution of weak base with unknown concentration. What would be a good acid with which to titrate the weakly alkaline solution, in order to determine its concentration?

Possible Answers:

Sodium hydroxide

Citric acid

Stearic acid

Nitric acid

Hydrofluoric acid

Correct answer:

Nitric acid

Explanation:

When you titrate a weak base, you want to titrate it with a strong acid. Hydrofluoric acid, citric acid, and stearic acid are all weak acids, and sodium hydroxide is a strong base. The best choice it nitric acid, a strong acid.

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Example Question #6 : Help With Titration Curves

You have a 500mL solution of a monoprotic acid with unknown concentration. You titrate it to completion with 36mL of 0.4M NaOH solution. What is $[HX]$ ?

Possible Answers:

$1.44*10^{-2}M$

$2.88\hspace{1 mm}M$

$5.56*10^{-2}M$

$2.88*10^{-2}M$

There is not enough information to solve

Correct answer:

$2.88*10^{-2}M$

Explanation:

If we are working with a monoprotic acid, our chemical equation is:

$NaOH+HX\rightarrow NaX+H_2O$

Now we will calculate the moles of $HX$ in our solution:

$36\hspace{1 mm}mL\times \frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times \frac{0.4\hspace{1 mm}moles\hspace{1 mm}NaOH}{1\hspace{1 mm}L}\times\frac{1\hspace{1 mm}moles\hspace{1 mm}HX}{1\hspace{1 mm}mole\hspace{1 mm}NaOH}=1.44\times 10^{-2}\hspace{1 mm}moles\hspace{1 mm}HX$

Now we will determine the concentration from the amount of moles and the volume

$\frac{1.44\times 10^{-2}\hspace{1 mm}moles\hspace{1 mm}HX}{500\hspace{1 mm}mL}\times\frac{1000\hspace{1 mm}mL}{1\hspace{1 mm}L}=2.88\times 10^{-2}\hspace{1 mm}M$

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Example Question #7 : Help With Titration Curves

0.458g of an unknown diprotic acid is dissolved in water. It is then titrated with 21.5ml of a 0.500M NaOH solution to reach the second equivalence point. Determine the molecular weight of this unknown acid.

Possible Answers:

$170.4\frac{g}{mol}$

$35.3\frac{g}{mol}$

$85.2\frac{g}{mol}$

$42.6\frac{g}{mol}$

$21.3\frac{g}{mol}$

Correct answer:

$85.2\frac{g}{mol}$

Explanation:

To find the molecular weight, we must determine the number of moles that correspond to the 0.458g sample. At the equivalence point, the moles of hydronium ions will equal the moles of hydroxide ions.

We need to use the molarity and volume of the NaOH that was added to find the number of moles of base added. This will tell us the moles of hydroxide ions.

$mol_{acid}=mol_{base}=M*V$

$(0.500 \frac{mol}{L})(0.0215L) = 0.01075 mol$

Now, since we are working with a diprotic acid, two moles of base would be required for every one mole of the acid. The moles of acid would be:

$0.01075mol\ NaOH*\frac{1mol\ \text{acid}}{2mol\ NaOH}=0.005375mol\ \text{acid}$

Now that we know the moles of acid in the sample, we can use the given sample mass to find the molecular weight.

$MW = \frac{0.458 g}{ 0.005375 mol} = 85.2 \frac{g}{mol}$

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High School Chemistry : Titrations

Study concepts, example questions & explanations for High School Chemistry

All High School Chemistry Resources

Example Questions

Example Question #3 : Titrations And Indicators

Example Question #6 : Help With Titration Curves

Example Question #7 : Help With Titration Curves

All High School Chemistry Resources

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