Titrations - High School Chemistry

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Question

A buffer using acetic acid (pKa=4.76) is titrated with NaOH. What is the pH at half the equivalence point?

Answer

The pH at half the equivalence point is equal to the pKa of the acid.

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Question

At what pH does the equivalence point lie for a strong-acid / strong-base titration?

Answer

The equivalence point for a strong-acid / strong-base titration will be at neutral pH, 7. This is because each equivalent of the acid will neutralize each equivalent of the base, and you will be left with a neutral solution.

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Question

At what pH does the equivalence point lie for a strong-acid / weak-base titration?

Answer

The equivalence point for a strong-acid / weak-base titration will be at a slightly acidic pH. This is because the acid is stronger and dissociates to a greater degree, while the base is not quite as strong, so doesn't dissociate to a large enough extent to neutralize each equivalent of the acid.

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Question

At what pH does the equivalence point lie for a weak acid-strong base titration?

Answer

The equivalence point for a weak-acid / strong-base titration will be at a slightly basic pH. This is because the base is stronger and dissociates to a greater degree, while the acid is not quite and strong and doesn't dissociate to a large enough extent to neutralize each equivalent of the base.

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Question

Where does the flattest region of a titration curve of the titration of a weak acid with a strong base occur?

Answer

In this question, titration curve would graph the pH of acid solution versus the amount of base added. Since the base is strong and the acid is weak, we can conclude that the pH will be slightly greater than 7 at the equivalence point. The equivalence point is found in the steepest region of the curve.

The half-equivalence point is the flattest region of the titration curve and is most resistant to changes in pH. This corresponds to the pKa of the acid. Within this region, adding base (changing the x-value) results in very little deviation in the pH (the y-value). This region is also the buffer region for the given acid.

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Question

You are given 500 mL of a HCl solution of unknown concentration and you titrate is with 0.0540 M NaOH. It takes 32.1 mL of the NaOH solution to reach your end point. What is $[HCl]$ of your original solution?

Answer

First, let us write out the reaction that occurs:

$HCl_{(aq)}+NaOH_{(aq)}\rightarrow NaCl_{(aq)}+H_2O_{(l)}$

$32.1\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{0.0540\hspace{1 mm}moles\hspace{1 mm}NaOH}{1\hspace{1 mm}L}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}HCl}{1\hspace{1 mm}mole\hspace{1 mm}NaOH}=1.73\times 10^{-3}\hspace{1 mm}moles\hspace{1 mm}HCl$

$\frac{1.73\times 10^{-3}\hspace{1 mm}moles\hspace{1 mm}HCl}{500\hspace{1 mm}mL}\times\frac{1000\hspace{1 mm}mL}{1\hspace{1 mm}L}=3.47\times 10^{-3}\hspace{1 mm}M$

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Question

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 $[HX]$ ?

Answer

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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Question

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?

Answer

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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Question

A titration is a drop-by-drop mixing of an acid and a base in order to determine the concentration of an unknown solution, via addition of a solution with known concentration. A titration curve can be graphed showing the relationship between the mixture pH and the amount of known solution added.

What would the titration curve look like for a strong base being titrated with a strong acid?

Answer

There are two things to consider here.

1. Since the solution is originally a strong base, the pH will be originally elevated. As a strong acid is added to the solution, the pH will decrease. As a result, the titration curve will be decreasing as the volume of titrant increases.

2. The titration curve will never be a straight line. Eventually, the strong acid will be much larger in volume than the original base; however, the pH will eventually even out at the pH of the added titrant.

Since we are titrating a strong base with a strong acid, the titration curve will be represented by a decreasing sigmoidal curve.

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Question

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.

Answer

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.

$mol_{acid}=mol_{base}=MV$

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

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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Question

What is the concentration if 40 g NaOH is dissolved in 1000 g of water

Answer

first of all, M = molar; m = molal- M = mol solute/ L of solution; m = mol solute/ kg solvent

you have 40 g NaOH * 1 mol/40 g = 1 mol

1000 g of water is equivalent to 1 L

1 mol/L = 1M

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Question

80.0g NaOH is put into 50000 mL water. What is the molarity of the resulting solution?

Answer

Molarity = mol solute / L soution

mol solute = 80 g NaOH * 1 mol / 40 g = 2 mol

L solution = 50000 mL water * 1 L/1000 mL = 50 L

2 mol / 50 L =

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Question

What is the molality of a solution created by mixing 4.3g $NaCl$ into 43g of water?

Answer

Molality can be defined:

$Molality=\hspace{1 mm}\frac{mol\hspace{1 mm}solute}{kg\hspace{1 mm}solvent}$

It is slightly different from Molarity and has different uses.

$Molality=\hspace{1 mm}\frac{4.3\hspace{1 mm}g\hspace{1 mm}NaCl}{43\hspace{1 mm}g\hspace{1 mm}H_20}\times\frac{1000\hspace{1 mm}g\hspace{1 mm}H_2O}{1\hspace{1 mm}kg\hspace{1 mm}H_2O}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}NaCl}{58.44\hspace{1 mm}g\hspace{1 mm}NaCl}=1.71\hspace{1 mm}m$

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Question

How much solid NaOH must be dissolved to make 740mL of a 0.32M solution?

Answer

This problem can be solved by stoichiometry. Remember that 0.32M gives us the moles of NaOH per liter, and solve for the number of moles per 0.740L.

$740\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{0.32\hspace{1 mm}moles\hspace{1 mm}NaOH}{1\hspace{1 mm}L}\times\frac{40\hspace{1 mm}g\hspace{1 mm}NaOH}{1\hspace{1 mm}mole\hspace{1 mm}NaOH}=9.47\hspace{1 mm}g\hspace{1 mm}NaOH$

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Question

A 100mL solution is composed of 25% ethanol by volume and water. What is the mass of the solution?

$\rho_{ethanol}=0.789\frac{g}{mL}$

Answer

First we determine the mass of the ethanol in solution using its density. Using the percent by volume of ethanol, we know that there are 25mL of ethanol in a 100mL solution. The remaining 75mL are water.

$25\hspace{ 1 mm}mL\hspace{ 1 mm}EtOH\times\frac{0.789\hspace{ 1 mm}g}{1\hspace{ 1 mm}mL}=19.7\hspace{ 1 mm}g\hspace{ 1 mm}EtOH$

Since the density of water is 1g/mL, we know that the mass of 75mL of water is 75g. The total mass is the sum of the ethanol and the water.

$19.7\hspace{ 1 mm}g+75.0\hspace{ 1 mm}g= 94.7\hspace{ 1 mm}g$

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Question

How many liters of 10M HCl are needed to make 4L of 0.4M solution?

Answer

A simple calculation can be done to perform any solution dilution problem. We know our equation $molarity = \frac{moles}{volume}$ .

We can rewrite this as .

Using this formula, we take the old solution and set it equal to the new solution.

$V = \frac{1.6mol}{10M}=0.16L$

We need 0.16 liters of our 10 molar solution.

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Question

Find the mass of in 10L of water if it is a 2m solution.

Answer

Molality is grams of solute per kilogram of solvent.

$m=\frac{mol}{kg}\rightarrow mol=(m)(kg)$

Water has a density of one gram per mililiter, so one liter of water equal to one kilogram. If we have a 2m solution, that means we have two moles of per kilogram of water.

$10L\frac{1kg}{1L}=10kg$

has a molecular weight of $\small 56\frac{g}{mol }$ .

$20mol\frac{56g}{1mol}=1120g\ KOH$

This gives us of .

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Question

Which solution has the highest molarity?

Answer

This question requires us to calculate molarity for each answer choice. It is important to add everything correctly and be careful with more complex compounds.

Molarity is simply moles of solute over liters of solution. The correct answer, after trying each, is the answer with lead (II) nitrate, as it gets us a molarity of 2.

$Pb(NO_3)_2:\ \frac{486.5g\frac{1mol}{331g}}{0.75L}=1.96M$

$NaCl:\ \frac{174g\frac{1mol}{58.5g}}{2.0L}=1.49M$

$MgSO_4:\ \frac{240g\frac{1mol}{120.3g}}{3.0L}=0.67M$

$NaNO_3:\ \frac{85g\frac{1mol}{85g}}{2.0L}=0.50M$

$CH_3CH_2OH:\ \frac{138g*\frac{1mol}{46g}}{3.0L}=1.0M$

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Question

Which of the following aqueous solutions is the most concentrated?

$\small \rho_{water}=1 \frac{g}{mL}$

Answer

In order to answer this question, it helps to know that 1 kilogram of water is equal to 1 liter of water, due to its density. Two of the above options refer to a 1m solution of hydrochloric acid. The other is a 1M solution.

$1M\ HCl=\frac{1mol\ HCl}{1L}$

$1m\ HCl=\frac{1mol\ HCl}{1kg\ water}=\frac{1mol\ HCl}{1L\ water}$

$36.5g\ \text{HCl in}\ 1kg\ \text{water}=\frac{1mol\ HCl}{1kg\ water}=\frac{1mol\ HCl}{1L\ water}$

All three of the options have the same amount of hydrochloric acid (one mole). For molarity, the hydrochloric acid is diluted with water until one liter of solution is created. For molality, one mole of HCl is added to one kilogram of water. Since one kilogram of water is one liter, this becomes the same concentration.

One a very small level, the 1M HCl solution will be slightly more concentrated. Creating a molal solution does not take into account the volume of the solute. If, for example, 100 cubic centimeters of HCl were added to one kilogram of water, the resulting volume would be more than one liter, making the concentration slightly less than 1M. This discrepancy is usually not accounted for in basic chemistry, but you should be familiar with the concept.

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Question

10mL of a solution of of unknown concentration mixed with 34mL of $10^{-4}M$ produced a solution with a pH of 7. What is the concentration of the solution?

Answer

For this question use the following formula:

is the number of acidic hydrogens on the acid, is the molarity of the acid, is the volume of the acid, is the number of basic hydroxides on the base, is the molarity of the base, is the volume of the base

Rearrange the equation for the molarity of the base:

$\frac{N_A M_A V_A }{N_B V_B}= M_B$

Plug in known values and solve.

$M_B=3.4\cdot 10^{-4}M$

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