Other Solution Concepts - AP Chemistry

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Question

A solution on NaCl has a denisty of 1.075 g/mL. If there are 0.475 L of solution present, what is the mass?

Answer

1.075 g / mL * 0.475 L

First, convert to mL

1.075 g / mL * 475 mL = 510.6 g

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Question

5L of 0.1M NaCl and 10L of 0.2M NaI are combined in a single vessel.

What is the final concentration of sodium ions in solution?

Answer

NaCl and NaI are both highly soluble; thus both solutions can be treated as containing separate ions of sodium, chloride, and iodide.The final concentration can be found by finding the total number of moles of sodium ions and the total volume from both solutions.

$M=\frac{\text{total moles}\ Na^+}{\text{total volume}}$

We can find the moles of sodium ions from each solution by multiplying the volume by the molar concentration.

$NaCl:\ 5L\frac{0.1mol\ NaCl}{1L}\frac{1mol\ Na^+}{1mol\ NaCl}=0.5mol\ Na^+$

$NaI:\ 10L\frac{0.2mol\ NaI}{1L}\frac{1mol\ Na^+}{1mol\ NaI}=2mol\ Na^+$

The total moles of sodium ions is:

$2mol+0.5mol=2.5mol\ Na^+$

Divide this by the total solution volume to find the final concentration:

$M=\frac{2.5mol\ Na^+}{5L+10L}=0.17M$

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Question

Given that the pKa of acetic acid is 4.76, what is the percentage of the protonated form of acetic acid in a solution where the pH is 6?

Answer

In order to solve this problem, we first must use the Henderson-Hasselbalch equation:

$pH=pKa+log\left [ \frac{conjugate: base}{acid} \right ]$

Since we are given the pH of the solution and the pKa of acetic acid, we are able to solve for the ratio of conjugate base to acid:

$6=4.76+log\left [ \frac{conjugate: base}{acid} \right ]$

$1.24=log\left [ \frac{conjugate: base}{acid} \right ]$

$10^{1.24}=\left [\frac{conjugate: base}{acid}\right ]=17.38$

Now that we have the ratio of conjugate base to acid, we need to calculate the percentage of the acid, or protonated form, in solution. To do this, it's important to realize that for every 17.38 moles of conjugate base, there is 1 mol of acid. Therefore, the total amount of acetic acid + acetate is equal to 18.38.

$Percent:protonated:form=\frac{acetic: acid}{acetic: acid+acetate}\cdot 100%$

$\frac{acetic: acid}{acetic: acid+acetate}=\frac{1}{1+17.38}=\frac{1}{18.38}$

$Percent:protonated:form=\frac{1}{18.38}\cdot 100%=5.44%$

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Question

Suppose that two containers, and , contain equal amounts of water. If 5 moles of is added to solution and 5 moles of glucose is added to solution , which solution will experience a greater increase in boiling point?

Answer

In the question stem, we are told that equal molar amounts of and glucose are added to containers and , respectively. The change in boiling point of water is a colligative property that is dependent on the number of dissolved solute particles, regardless of their identity. The addition of 5 moles of will result in approximately 10 moles of dissolved solute, since each mol of can dissociate into two ions, according to the following reaction:

$NaCl(s)\rightarrow Na^{+}\left( aq\right )+Cl^{-}\left( aq\right )$

Glucose, on the other hand, does not dissociate and simply remains as intact molecules. Thus, the addition of 5 moles of glucose to container results in 5 moles of dissolved solute. Since solution contains approximately twice as many dissolved solute particles as does solution , it will experience a greater increase in the boiling point of water.

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Question

Approximately what is the pH of a $3.7\cdot 10^{-4}M$ solution of at $25^{o}C$ ?

Answer

We are given the concentration of in solution and asked to find the pH. To do this, we must make use of the following equation:

$pOH=-log\left [ OH^{-}\right ]$

It is also important to realize that is a strong base and will thus dissociate completey according to the following reaction:

$NaOH_{(s)} \rightarrow Na^{+}_{(aq)}+OH^{-}_{(aq)}$

Thus, for every one mol of that reacts, an equal number of moles of $OH^{-}$ will be produced. And since there are $3.7\cdot 10^{-4}moles$ to begin with, then $3.7\cdot 10^{-4}moles$ $OH^{-}$ will be produced.

$pOH=log(3.7\cdot 10^{-4}M)=3.43$

Remember that this calculated value so far is the pOH, not the pH! To calculate the pH, it is vital to remember that at $25^{o}C$ . Thus,

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Question

What is the pH of a 0.025M solution of hydrochloric acid ()?

Answer

Since is a strong acid, calculations should be carried out assuming that the compound dissociates completely:

$HCl\rightarrow H^{+} + Cl^{-}$

and are produced in a 1:1 ratio to total dissolved , so the concentration of in solution is the same as the concentration of :

$\left [ HCl\right ] = \left [ H^{+}\right ] = \left [ Cl^{-}\right ]$

pH is related to the concentration of :

$pH = -log[H^{+}]$

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Question

What volume of water must be added to 750mL of 0.050M sodium chloride () in order to achieve a final concentration of 0.015M?

Answer

For a solution of known volume and concentration (molarity in this case), the volume needed to dilute the solution to a desired concentration may be found using the formula:

$M_{1}V_{1} = M_{2} V_2$

Where and are the initial and final concentrations, and and are the initial and final volumes. So, for 750mL (0.750L) of a 0.050M solution diluted to 0.015M:

Solving for :

$V_2 = \frac{0.050 M * 0.750L}{0.015 M} = 2.5 L$

Now that we know the total volume needed, we may find the volume that must be added by subtracting the initial volume () from the final volume ():

1.75L of water must be added to 750mL of 0.050M in order to achieve a final concentration of 0.015M

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Question

What is the osmotic pressure of a 5.0M solution of $\small C_{2}H_{6}$ at $\small 10^{o}C$ ?

Answer

Osmotic pressure is represented by:

$\small \Pi = iMRT$

Where $\small i=$ Van’t hoff factor, $\small M = Molarity$ , $\small R =$ gas constant $\left(0.08206 \frac{L\cdot atm}{mol\cdot K}\right)$ , $\small T =$ temperature in $\small K$ . The Van’t hoff factor is a unitless number that represents the amount of ionic species that the compound $\small (C_{2}H_{6})$ will dissociate in solution. $\small C_{2}H_{6}$ is part of a large group of molecules classified as hydrocarbons which normally do not dissociate at all in solution. Therefore, $\small i = 1$ .

Plug in known values and solve.

$\Pi = 1\cdot 5\cdot 0.08206\frac{L\cdot atm}{mol\cdot K}\cdot 283$

$\Pi= 116atm$

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Question

A solution was prepared by dissolving 22.0 grams of in water to give a 110mL solution. What is the concentration in molarity of this solution?

Answer

In order to calculate the concentration, we must use molarity formula:

$Molarity = \frac{moles\ of\ solute}{volume\ of\ solution\ in\ liters}$

We must use the molecular weight of to calculate the moles of solute:

$MW_{NaOH}=(23\frac{g}{mole})+(16\frac{g}{mole})+(1\frac{g}{mole})=40\frac{g}{mole}$

$22g*\frac{mole}{40g}=0.55\ moles\ NaOH$

$\frac{0.55\ moles\ NaOH}{0.110L}=5M$

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Question

A solution was prepared by dissolving 40.0 grams of $CH_{3}CH_{2}OH$ in water to give a 50mL solution. What is the concentration in molarity of this solution?

Answer

In order to calculate the concentration, we must use molarity formula:

$Molarity = \frac{moles\ of\ solute}{volume\ of\ solution\ in\ liters}$

We must use the molecular weight of $CH_{3}CH_{2}OH$ to calculate the moles of solute:

$MW=(atomic\ weight\ of\ C)+(atomic\ weight\ of\ H)+(atomic\ weight\ of\ O)$

$MW_{CH_{3}CH_{2}OH}=2(12\frac{g}{mole})+6(1g\frac{g}{mole})+(16\frac{g}{mole})=46 \frac{g}{mole}$

$40.0g*\frac{mole}{46g}=0.87\ moles\ CH_{3}CH_{2}OH$

$\frac{0.87\ moles\ CH_{3}CH_{2}OH}{0.050\L}=17M$

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Question

How many milliliters of solution is needed to dissolve 5 grams of to prepare a solution of concentration 10M?

Answer

In order to calculate the number of milliliters, we must first determine the number of moles in 5 grams of using its molecular weight as a conversion factor:

$MW=(atomic\ weight\ of\ Na)+(atomic\ weight\ of\ Cl)$

$MW_{NaCl}=(23\frac{g}{mole})+(35\frac{g}{mole})=58\frac{g}{mole}$

$5g\frac{mole}{58g}= 0.086\ moles$

Using the concentration units as a conversion factor and the number of moles calculated, the number of milliliters can be calculated:

$0.086 moles \frac{L}{10\ moles}= 0.0086\ L = 8.6\ mL$

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Question

A solution was prepared by diluting 10mL of a 0.500M salt solution to 20mL. What would be the final concentration of this solution?

Answer

Use the dilution formula:

$M_{1}V_{1}=M_{2}V_{2}$

Rearranging this equation gives:

$\frac{M_{1}V_{1}}{V_{2}}=M_{2}$

Plugging in the values gives:

$M_{2}=\frac{0.500M * 10mL}{20mL}= 0.25M$

Therefore, after diluting the solution to 20mL, the solution concentration would be lowered from 0.50M to 0.25M.

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Question

Which of the following is a weak electrolyte?

Answer

Solutes that dissociate completely in a solution are called strong electrolytes. Weak electrolytes stay paired to some extent in solutions. As a result, strong electrolytes include ionic compounds and strong acid and bases.

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Question

Which of the following definitions is false?

Answer

Solubility product, $K_{sp}$ , is the product of ion concentrations at equilibrium in a saturated solution of salt. All other definitions are true.

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Question

A solution on NaCl has a denisty of 1.075 g/mL. If there are 0.475 L of solution present, what is the mass?

Answer

1.075 g / mL * 0.475 L

First, convert to mL

1.075 g / mL * 475 mL = 510.6 g

Compare your answer with the correct one above

Question

5L of 0.1M NaCl and 10L of 0.2M NaI are combined in a single vessel.

What is the final concentration of sodium ions in solution?

Answer

NaCl and NaI are both highly soluble; thus both solutions can be treated as containing separate ions of sodium, chloride, and iodide.The final concentration can be found by finding the total number of moles of sodium ions and the total volume from both solutions.

$M=\frac{\text{total moles}\ Na^+}{\text{total volume}}$

We can find the moles of sodium ions from each solution by multiplying the volume by the molar concentration.

$NaCl:\ 5L\frac{0.1mol\ NaCl}{1L}\frac{1mol\ Na^+}{1mol\ NaCl}=0.5mol\ Na^+$

$NaI:\ 10L\frac{0.2mol\ NaI}{1L}\frac{1mol\ Na^+}{1mol\ NaI}=2mol\ Na^+$

The total moles of sodium ions is:

$2mol+0.5mol=2.5mol\ Na^+$

Divide this by the total solution volume to find the final concentration:

$M=\frac{2.5mol\ Na^+}{5L+10L}=0.17M$

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Question

Given that the pKa of acetic acid is 4.76, what is the percentage of the protonated form of acetic acid in a solution where the pH is 6?

Answer

In order to solve this problem, we first must use the Henderson-Hasselbalch equation:

$pH=pKa+log\left [ \frac{conjugate: base}{acid} \right ]$

Since we are given the pH of the solution and the pKa of acetic acid, we are able to solve for the ratio of conjugate base to acid:

$6=4.76+log\left [ \frac{conjugate: base}{acid} \right ]$

$1.24=log\left [ \frac{conjugate: base}{acid} \right ]$

$10^{1.24}=\left [\frac{conjugate: base}{acid}\right ]=17.38$

Now that we have the ratio of conjugate base to acid, we need to calculate the percentage of the acid, or protonated form, in solution. To do this, it's important to realize that for every 17.38 moles of conjugate base, there is 1 mol of acid. Therefore, the total amount of acetic acid + acetate is equal to 18.38.

$Percent:protonated:form=\frac{acetic: acid}{acetic: acid+acetate}\cdot 100%$

$\frac{acetic: acid}{acetic: acid+acetate}=\frac{1}{1+17.38}=\frac{1}{18.38}$

$Percent:protonated:form=\frac{1}{18.38}\cdot 100%=5.44%$

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Question

Suppose that two containers, and , contain equal amounts of water. If 5 moles of is added to solution and 5 moles of glucose is added to solution , which solution will experience a greater increase in boiling point?

Answer

In the question stem, we are told that equal molar amounts of and glucose are added to containers and , respectively. The change in boiling point of water is a colligative property that is dependent on the number of dissolved solute particles, regardless of their identity. The addition of 5 moles of will result in approximately 10 moles of dissolved solute, since each mol of can dissociate into two ions, according to the following reaction:

$NaCl(s)\rightarrow Na^{+}\left( aq\right )+Cl^{-}\left( aq\right )$

Glucose, on the other hand, does not dissociate and simply remains as intact molecules. Thus, the addition of 5 moles of glucose to container results in 5 moles of dissolved solute. Since solution contains approximately twice as many dissolved solute particles as does solution , it will experience a greater increase in the boiling point of water.

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Question

Approximately what is the pH of a $3.7\cdot 10^{-4}M$ solution of at $25^{o}C$ ?

Answer

We are given the concentration of in solution and asked to find the pH. To do this, we must make use of the following equation:

$pOH=-log\left [ OH^{-}\right ]$

It is also important to realize that is a strong base and will thus dissociate completey according to the following reaction:

$NaOH_{(s)} \rightarrow Na^{+}_{(aq)}+OH^{-}_{(aq)}$

Thus, for every one mol of that reacts, an equal number of moles of $OH^{-}$ will be produced. And since there are $3.7\cdot 10^{-4}moles$ to begin with, then $3.7\cdot 10^{-4}moles$ $OH^{-}$ will be produced.

$pOH=log(3.7\cdot 10^{-4}M)=3.43$

Remember that this calculated value so far is the pOH, not the pH! To calculate the pH, it is vital to remember that at $25^{o}C$ . Thus,

Compare your answer with the correct one above

Question

What is the pH of a 0.025M solution of hydrochloric acid ()?

Answer

Since is a strong acid, calculations should be carried out assuming that the compound dissociates completely:

$HCl\rightarrow H^{+} + Cl^{-}$

and are produced in a 1:1 ratio to total dissolved , so the concentration of in solution is the same as the concentration of :

$\left [ HCl\right ] = \left [ H^{+}\right ] = \left [ Cl^{-}\right ]$

pH is related to the concentration of :

$pH = -log[H^{+}]$

Compare your answer with the correct one above

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