Analyzing Reactions - GRE Subject Test: Chemistry

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Question

47.0g of nitrous acid, HNO2, is added to 4L of water. What is the resulting pH? $\dpi{100} \small \left (K_{a}=4.1\times 10^{-4} \right )$

Answer

HNO2 is a weak acid; it will not fully dissociate, so we need to use the HA → H+ + A– reaction, with $\dpi{100} \small K_{a}=\frac{\left [ products \right ]}{\left [ reactants \right ]}=\left \frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}=4.1\times 10^{-4}$ .

47.0g HNO2 is equal to 1mol. 1mol into 4L gives a concentration of 0.25M when the acid is first dissolved; however, we want the pH at equilibrium, not at the initial state. As the acid dissolves, we know \[HNO2\] will decrease to become ions, but we don't know by how much so we indicate the decrease as "x". As HNO2 dissolves by a factor of x, the ion concentrations will increase by x.

HNO2 → H+ + NO2–

Initial 0.25M 0 0

Equilibrium 0.25 – x x x

Now, we can fill in our equation: $\dpi{100} \small K_{a}=\left \frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}=\frac{\left ( x \right )\left ( x \right )}{0.25-x}$ .

Since x is very small, we can ignore it in the denominator: $\dpi{100} \small K_{a}=\left \frac{\left ( x \right )\left ( x \right )}{\left (0.25 \right )}=4.1\times 10^{-4}$

(they expect you to do this on the MCAT; you will never have to solve with x in the denominator on the exam!)

Solve for x, and you find $x=1\times10^{-2}$ . Looking at our table, we know that $\dpi{100} \small x=\left [ H^{+} \right ]$

Now we can solve for pH: $pH=-log[H^+]=-log(1\times10^{-2})=2.0$

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Question

NaOH is added to a 500mL of 2M acetic acid. If the pKa value of acetic acid is approximately 4.8, what volume of 2M NaOH must be added so that the pH of the solution is 4.8?

Answer

To solve this question you need to think about the chemical reaction occurring.

$CH_3COOH+NaOH\rightleftharpoons CH_3COO^-+H_2O+Na^+$

We can ignore water and sodium ions for the sake of this question. The reactants exist in a 1:1 ratio, so that for every mol of NaOH we add, we lose one mol of acetic acid and gain one mol of acetate. We can determine the moles of acetic acid by using M = mol/L, which gives us mol = ML = (2M) * (0.5L) = 1mol acetic acid. If we use the Hendersen Hasselbach equation we can see that the pH equals the pKa when the concentration of conjugate base (acetate) equals the concentration of acid.

$pH=pK_a+log\frac{[base]}{[acid]}$

If we have 1mol of acetic acid and add 0.5mol of NaOH, we will lose 0.5mol of acetic acid and gain 0.5mol of acetate. We will then be at a point where acetic acid equals acetate. This is summarized in the ICE table below. Now we know the moles of NaOH (0.5 moles) and the concentration (2M) so we can find the volume by doing M = mol/L.

L = mol/M = (0.5mol)/(2M) = 0.25L

| | Acetic acid | NaOH | Acetate | | | -------------- | --------- | --------- | -------- | | I | 1 mol | 0.5 mol | 0 mol | | C | -0.5 mol | -0.5 mol | +0.5 mol | | E | 0.5 mol | 0 mol | 0.5 mol |

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Question

is a diprotic acid. Which of the following statements is true about its second half equivalence point?

Answer

H2CO3 will have two equivalence points and two half equivalence points, one set corresponding to each of its two protons.

After the first equivalence point, all of the acid is in the form HCO3-. When an acid is at a half equivalence point, the acid's concentration will be equal to the concentration of the conjugate base. For the second half equivalence point, the acid is in the HCO3- form, and the conjugate base is CO32-. As a result, at the second half equivalence point, the concentrations of HCO3- and CO32- will be equal.

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Question

The Ka for HCN is $\small 6.2*10^{-10}$ .

If there is a solution of 2M HCN, what concentration of NaCN is needed in order for the pH to be 9.2?

Answer

To answer this question, we need to be able to compare the concentrations of acid and conjugate base in the solution with the pH. The Henderson-Hasselbach equation is used to compare these values, and is written as: $\small pH = pK_{a} + log\frac{[A^{-}]}{[HA]}$ . In this question, this equation can be written with the given values of Ka and pH.

$9.2=-log(6.2*10^{-10})+log\frac{[CN^-]}{[HCN]}$

Since we know the Ka of HCN, we can derive the pKa, which turns out to be 9.2.

$pH=9.2+log\frac{[CN^-]}{[HCN]}=9.2$

As a result, we want to see to it that the amount of conjugate base is equal to the concentration of acid, so that $\small log\frac{[A^{-}]}{[HA]} =0$ . Because log(1) = 0, we want to see to it that the concentrations of the acid and the conjugate base are equal to one another. We know from the question that \[HCN\] = 2M.

$pH=9.2+log\frac{[CN^-]}{[2]}=9.2$

$\frac{[CN^-]}{[2]}=1$

As a result, a concentration of 2M NaCN will allow the pH of the solution to be 9.2.

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Question

Consider the titration curve, in which an acid is treated with aqueous 0.5M NaOH.

The pH at the equivalence point was found to be 8.9. Approximately how many milliliters of 0.5M NaOH were needed to reach equivalence?

Answer

For any titration curve the equivalence point corresponds to the steepest part of the curve. In this case, the question indicates that the pH at equivalence was 8.9, which would correspond to a volume between 20.0 and 24.0 on the x-axis of the graph. This leads to our approximate answer of 21.8mL.

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Question

Consider the titration curve, in which an acid is treated with aqueous 0.5M NaOH.

What is the approximate pKa of the starting acid?

Answer

The pKa corresponds to the pH when half of the acid has been neutralized. In this case, that would correspond to when half of the 0.5M NaOH has been added. We can see in the curve that the equivalence point is at around 22mL of NaOH, as this is the steepest change in pH. The half equivalence point will be found at approximately 11mL of NaOH addition. On the graph, 11mL of NaOH results in a pH of approximately 5.0, giving us our pKa value.

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Question

Calculate the concentration of hydrogen ions in the following acetic acid solution.

$\frac{[CH_3COOH]}{[CH_3COO^-]} = 0.5$

Answer

To answer this question you need to use the Henderson-Hasselbalch equation:

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

The ratio given in the question is $\frac{[CH_3COOH]}{[CH_3COO^-]}$ , or $\frac{[acid]}{[conjugate : base]}$ .

To use the correct ratio for the Henderson-Hasselbalch equation, we need to convert this ratio to its reciprocal:

$\frac{CH_3COO^-}{CH_3COOH} = \frac{1}{0.5} = 2$

Plugging the given values into the equation gives us:

The question is asking for the concentration of hydrogen ions. To solve for this we have to use the definition of pH.

Solving for the concentration of hydrogen ions gives us:

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Question

Which of the following is true regarding the Henderson-Hasselbalch equation?

I. The pH of the solution is always greater than the pKa of the solution

II. As the ratio of conjugate base to acid increases, the pH increases

III. The hydrogen ion concentration can never equal the acid dissociation constant

Answer

The Henderson-Hasselbalch equation is a tool that allows us to calculate the pH of an acid solution using the pKa of the acid and the relative concentrations of the acid and its conjugate base. It is defined as:

$pH = pKa + log\frac{[A-]}{[HA]}$

By looking at the equation we can determine that if the ratio inside the logarithm is greater than 1, then the pH of the solution will be greater than the pKa; however, if the ratio is less than 1 (meaning, if the concentration of the acid is greater than the concentration of conjugate base), then the pH will be less than the pKa. Statement I is false.

Increasing the ratio of to will increase the logarithm, and subsequently the pH of the solution. This makes sense because you will have more conjugate base than acid, thereby making the solution more alkaline and increasing the pH. Statement II is true.

pH and pKa are defined as follows:

If we have the same concentration of hydrogen ions as the acid dissociation constant (), then the pH will equal the pKa. According to the Henderson-Hasselbalch equation, the pH equals the pKa if the concentration of the conjugate base equals the concentration of acid; therefore, it is possible for the hydrogen ion concentration to equal the acid dissociation constant. Statement III is false.

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Question

A researcher prepares two solutions. Solution A contains an unknown acid, HA, and solution B contains an unknown acid, HB. The researcher performs several tests and collects the following data.

1. Both solutions contain weak acids

2. $\frac{[A^-]}{[HA]} = 1000$

3.

4. $\frac{[B^-]}{[HB]} = 100$

5.

What can you conclude about these two solutions?

Answer

The Henderson-Hasselbalch equation states that:

$pH = pKa + log\frac{[A^-]}{[HA]}$

The question gives us information regarding the ratio of conjugate base to acid AND the pH for each acidic solution. Using this information, we can solve for the pKa values of both solutions.

$pKa = pH - log\frac{[A^-]}{[HA]}$

$\text{pKa of solution A }= 5.60 - log (1000) = 2.6$

$\text{pKa of solution B }= 4.60 - log (100) = 2.6$

The pKa values of both solutions are the same. This means that both solution contains the same acid; therefore, the identity of HA is the same as the identity of HB.

The hydrogen ion concentration of solution A is lower than that of solution B because the pH of solution A is greater. Acidity, or strength, of an acid is determined by the pKa. Since we have the same pKa for both acids, HA and HB will have the same acidity. Acid dissociation constant, Ka, is defined as:

Acid dissociation constant only depends on the pKa; therefore, the Ka for both acids is the same.

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Question

What is the of a 0.12M solution of that has a pH of 5.1?

Answer

We can calculate the hydrogen ion concentration by using the following equation:

$[H^{+}]=10^{-pH}$

Plug in the given value for pH and solve:

$[H^{+}]=10^{-5.1}=7.94*10^{-6} M$

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Question

Based on the equilibrium above, what does $NH_{4}^{+}$ act as?

Answer

An acid is a substance that can donates a proton. The conjugate acid of a base is formed when the base accepts a proton. In this case, $NH_{4}^{+}$ is the conjugate acid to the base $NH_{3}$ . This is because $NH_{3}$ accepts a hydrogen ion from the water molecule to form $NH_{4}^{+}$ , the conjugate acid.

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Question

Given the following equilibria, what is the hydroxide ion concentration of a solution containing $0.06\textup{ M}$ $NH_{3}$ ?

$NH_{3}+H_{2}O\rightleftharpoons\ NH_{4}^{+}+OH^{-}$

The $K_{b}$ for $NH_{3}$ is $1.75\times10^{-5}$

Answer

Below is the acid-base equilibria of $NH_{3}$ in an aqueous solution:

$NH_{3}+H_{2}O\rightleftharpoons\ NH_{4}^{+}+OH^{-}$

A solution of ammonia ( $NH_{3}$ ) is basic based on the chemical equation given. The $K_{b}$ for this reaction is:

$Kb= \frac{[NH_{}4]^{+}][OH^{-}]}{[NH_{3}]}=1.75\times10^{-5}$

Based on the chemical equation, $[OH^{-}]=[NH_{4}^{+}]$ .

We can make the concentration of these species equal to :

$[OH^{-}]=[NH_{4}^{+}]=X$

The $NH_{3}$ concentration is equal to:

$NH_{3}=0.06-[OH^{-}]$

$NH_{3}$ has a low $K_{b}$ so we can assume the following:

$NH_{3}=0.06-[OH^{-}]\approx0.06$

Plugging the values into the base dissociation constant equation gives:

$Kb= \frac{[NH_{}4]^{+}][OH^{-}]}{[NH_{3}]}=\frac{X^{2}}{0.06}=1.75\times10^{-5}$

Solve for :

$X^{2}= (0.06)\times(1.75\times10^{-5})=1.05\times10^{-6}$

$X=\sqrt{1.05\times 10^{-6}}=0.0010\ M$

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Question

Given the following equilibria, what is the hydrogen ion concentration of a solution containing $0.01\textup{ M}$ $C_{6}H_{5}COOH$ (benzoic acid)?

$C_{6}H_{5}COOH(aq)\rightleftharpoons\ C_{6}H_{5}COO^{-}(aq)+H^{+}(aq)$

The $K_{a}$ for $NH_{3}$ is $6.3\times10^{-5}$ .

Answer

Below is the acid-base equilibria of $C_{6}H_{5}COOH$ in an aqueous solution:

$C_{6}H_{5}COOH(aq)\rightleftharpoons\ C_{6}H_{5}COO^{-}(aq)+H^{+}(aq)$

A solution of benzoic acid ( $C_{6}H_{5}COOH$ ) is acidic based on the chemical equation given. The $K_{a}$ for this reaction is:

$K_{a}= \frac{[C_{6}H_{5}COO^{-}][H^{+}]}{[C_{6}H_{5}COOH]}=6.3\times10^{-5}$

Based on the chemical equation, $[H^{+}]=[C_{6}H_{5}COO^{-}]$ .

We can make the concentration of these species equal to :

$[H^{+}]=[C_{6}H_{5}COO^{-}]=X$

The $C_{6}H_{5}COOH$ concentration is equal to:

$[C_{6}H_{5}COOH]=0.01-[H^{+}]$

$C_{6}H_{5}COOH$ has a low $K_{a}$ so we can assume the following:

$C_{6}H_{5}COOH=0.01-[H^{+}]\approx0.01$

Plugging the values into the base dissociation constant equation gives:

$Kb= \frac{[C_{6}H_{5}COO^{-}][H^{+}]}{[C_{6}H_{5}COOH]}=\frac{X^{2}}{0.01}=6.3\times10^{-5}$

Solve for :

$X^{2}= (0.01)\times(6.3\times10^{-5})=6.3\times10^{-7}$

$X=\sqrt{6.3\times10^{-7}}=7.9\times10^{-4}\textup{ M}$

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Question

Given the following equilibria, what is the hydronium ion concentration of a solution containing $0.025\textup{ M}$ ?

$HOCl(aq)+H_{2}O\rightleftharpoons\ OCl^{-}(aq)+H_{3}O^{+}(aq)$

The $K_{a}$ for is $3.0\times10^{-8}$ .

Answer

Below is the acid-base equilibria of in an aqueous solution:

$HOCl(aq)+H_{2}O_{(l)}\rightleftharpoons\ OCl^{-}(aq)+H_{3}O^{+}(aq)$

A solution of is acidic based on the chemical equation given. The Ka for this reaction is:

$K_{a}= \frac{[OCl^{-}][H_{3}O^{+}]}{[HOCl]}=3.0\times10^{-8}$

Based on the chemical equation, $[H_{3}O^{+}]=[OCl^{-}]$ .

We can make the concentration of these species equal to :

$[H_{3}O^{+}]=[OCl^{-}]=X$

The concentration is equal to:

$[HOCl]=0.025-[H^{+}]$

has a low $K_{a}$ so we can assume the following:

$HOCl=0.025-[H_{3}O^{+}]\approx0.025$

Plugging the values into the base dissociation constant equation gives:

$K_{a}= \frac{[OCl^{-}][H_{3}O^{+}]}{[HOCl]}=\frac{X^{2}}{0.01}=3.0\times10^{-8}$

Solve for :

$X^{2}= (0.025)\times(3.0\times10^{-8})=7.5\times10^{-10}$

$X=\sqrt{7.5\times10^{-10}}=2.7\times10^{-5}\textup{ M}$

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Question

$2NO_{(g)}+2H_{2 (g)}\leftrightharpoons N_{2 (g)}+2H_{2}O_{(g)}$

In the above reaction, by what factor would the reaction quotient change if the concentration of $\small NO$ were doubled?

Answer

For a general chemical equation $\alpha A + \beta B \rightleftharpoons \gamma C + \delta D$ , where A, B, C, and D are elements and the Greek letters are their coefficients, we have the reaction quotient equation:

$Q=\frac{[C]^{\gamma}[D]^{\delta}}{[A]^{\alpha}[B]^{\beta}}$

We can find the reaction quotient equation for our reaction by substituting the variables.

$2NO_{(g)}+2H_{2 (g)}\leftrightharpoons N_{2 (g)}+2H_{2}O_{(g)}$

$Q=\frac{[N_{2}][H_{2}O]^{2}}{[NO]^{2}{[H_{2}]}^{2}}$

Notice that the concentration of $\small NO$ is in the denominator and is squared, so doubling the concentration of $\small NO$ changes the reaction quotient by a factor of one-fourth.

$Q_2=\frac{[N_{2}][H_{2}O]^{2}}{[2NO]^{2}{[H_{2}]}^{2}}=\frac{[N_{2}][H_{2}O]^{2}}{4[NO]^{2}{[H_{2}]}^{2}}$

$Q_2=\frac{1}{4}K_c$

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Question

250mL of 2N is added to 100mL of 5N . An indicator in the solution is known to be yellow at any pH greater than 8.3 and green at any pH less than 8.3. Which of the following best describes the solution once it reaches equilibrium?

Answer

Each of these compounds requires one equivalent of H+ is added to 100mL of 5N NH3 mol, so for HClO4, 2N = 2M, and for NH3, 5N = 5M.

Using the concentrations and volumes, we can find the moles, finding $\dpi{100} \small 0.25\times 2=0.5mol$ HClO4 and $\dpi{100} \small 0.10\times 5=0.5mol$ NH3.

In this case, equivalents of acid are equal to the equivalents of base, meaning that we are at the equivalence point in a titration. HClO4 is a strong acid, and NH3 is a weak base.

Thus, the acid will fully dissociate, while the base will not, resulting in a greater concentration of H+ than OH– in the solution. This means the resulting solution will be acidic. We know that the indicator changes from yellow to green at 8.3, which is a basic pH. Our initial solution is basic, and we must pass through the pH of 8.3 to reach our final acidic solution, with pH < 8.3, meaning that the indicator must change from yellow to green during the reaction. This gives out final answer that the solution will be acidic and green.

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Question

Given the unbalanced equation below, how many grams of carbon dioxide will be produced from one mole of glucose and three moles of oxygen?

$C_6H_{12}O_6+O_2\rightarrow CO_2+H_2O$

Answer

The first step to solve will be to balance the chemical reaction:

$C_6H_{12}O_6+6O_2\rightarrow 6CO_2+6H_2O$

We see that we see that for every one mole of glucose used, six moles of carbon dioxide will be made. Similarly, for every six moles of oxygen used, six moles of carbon dioxide will be formed. For the reaction to carry out to completion, however, there must exist six moles of oxygen for every one mole of glucose. In the problem's circumstances, one of these compounds becomes the limiting reactant, in this case it is oxygen.

$1mol\ C_6H_{12}O_6\frac{6mol\ O_2}{1mol\ C_6H_{12}O_6}=6mol\ O_2$

We only have three moles of oxygen, but we would need six to react all the given glucose, making oxygen the limiting reagent. We need to find the carbon dioxide produced from the limited amount of oxygen present. Use the molar ratio between oxygen and carbon dioxide and the molar mass of carbon dioxide to solve.

$3mol\ O_2\frac{6mol\ CO_2}{6mol\ O_2}*\frac{44g\ CO_2}{1mol\ CO_2}=132g\ CO_2$

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Question

Which molecule would be considered a Lewis acid?

Answer

A lewis acid is an electron pair acceptor. $Zn^{2+}$ would be considered a lewis acid based on the definition. Because $Zn^{2+}$ is electron-deficient based on its oxidation number, it is able to accept an electron pair.

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Question

Which of the following would be considered a Lewis base?

Answer

A lewis base is an electron pair donor. $Br^{-}$ would be considered a lewis base based on the definition. Because $Br^{-}$ is electron rich based on its oxidation number, it is able to donate an electron pair to an electron deficient ion.

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Question

Which of the following types of reactions best describes the following reaction:

$Ce^{4+}+Fe^{2+}\rightarrow\ Ce^{3+}+Fe^{3+}$

Answer

A redox reaction is also known as an oxidation/reduction reaction. This type of reaction involves the transfer of electrons from on reactant to another. In this reaction, an electron is transferred from $Fe^{2+}$ to $Ce^{4+}$ forming the products $Ce^{3+}$ and $Fe^{3+}$ . The substance that gains an electron is referred to as being reduced. The substance that loses an electron is referred to as being oxidized.

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