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GRE Subject Test: Chemistry : General Chemistry

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Example Questions

Example Question #3 : Titrations

Screen shot 2015 10 25 at 8.56.18 pm

Based on the pH titration curve provided, how many acidic protons are in this acid?

Possible Answers:

Three

It cannot be determined from the graph

Four

One

Two

Correct answer:

Two

Explanation:

The equivalence point is the point in a titration at which the number of moles of titrant species equals the number of moles of analyte. For a polyprotic acid there are multiple equivalence points because there are more than one acidic proton in one molecule of the acid. A titration curve has as many equivalence points as the number of protons that may be neutralized by the interaction with a base. In this titration curve there are two equivalence points. The equivalence points occur when the graph has a very steep slope, and involve very large changes in pH with additions of small amounts of base. For this graph, the equivalence points occur at about pH 3.5 and 10.

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Example Question #112 : General Chemistry

What volume in liters of $0.1\textup{ M}$ NaOH is needed in the titration of $50.0\textup{ mL}$ of a $0.150\textup{ M}$ $CH_{3}COOH$ (acetic acid) solution to reach the equivalence point?

Possible Answers:

$0.075\textup{ L}$

$0.025\textup{ L}$

$1.1\textup{ L}$

$0.110\textup{ L}$

$2.20\textup{ L}$

Correct answer:

$0.075\textup{ L}$

Explanation:

The equivalence point in a titration is the point in which the number of moles of titrant equals to the number of moles of analyte:

$number\ of\ moles\ of\ acetic\ acid=number\ of\ moles\ NaOH$

We need to determine the number of moles of acetic acid we are dealing with:

$50.00\ mL=0.050\ L$

$\frac{0.150\ moles}{L}\times0.050L\ acetic\ acid=0.0075\ moles\ acetic\ acid$

Therefore, at the equivalence point there are also 0.0075 moles of NaOH in the solution. In order to determine the number of moles of NaOH in solution, we must use the concentration of NaOH as a conversion factor to determine the volume added to the acetic acid solution.

$0.0075\ moles\ NaOH\times \frac{L}{0.100\ moles}=0.075\ L\ NaOH$

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Example Question #11 : Titrations

A 0.4000g sample of a monoprotic acid was titrated against a 0.05M solution of NaOH . If the end point was reached after adding $30 mL\ NaOH$ , what is the molar mass of the weak acid?

Possible Answers:

$266.67\frac{g}{mol}$

$162.11\frac{g}{mol}$

$6.0\times10^{-4}\frac{g}{mol}$

$0.0038\frac{g}{mol}$

Correct answer:

$266.67\frac{g}{mol}$

Explanation:

A pH indicator added to a solution causes a color change which is an indication that a reaction has reached the equivalence point. The equivalence point is when the number of moles of titrant equals the number of moles of analyte. The end point is an approximation of this in a titration.

At the equivalence point:

$moles\ of\ NaOH=moles\ of\ acid$

The number of moles of NaOH at equivalence point is calculated as follows:

$0.030\ L\times \frac{0.05\ moles}{L}=0.0015\ moles\ NaOH$

$moles\ of\ acid= 0.0015\ moles$

Assuming we were trying to convert the number of grams of the monoprotic acid to moles, the equation would be set up as follows:

$grams\ of\ acid\ (g) \times\ molecular\ weight\left ( \frac{moles}{g} \right )=number\ of\ moles$

Let's plug the values we have into this equation and give the molar mass a value of :

$0.4000\ g\times \frac{1\ mole}{X}=0.0015\ moles$

Rearranging this equation to solve for the molar mass gives:

$X=\frac{0.4000\ g\times\ 1\ mole}{0.0015\ moles}=266.67\ grams$

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Example Question #21 : Acid Base Chemistry

What is the molarity of a HCl solution that has a pH of 5.0 ?

Possible Answers:

1.50 M

$5.5 * 10^{-1} M$

$1 * 10^{-5} M$

0.5M

3.50 M

Correct answer:

$1 * 10^{-5} M$

Explanation:

The pH of the solution is , therefore the concentration would be $1 * 10^{-5} M$ . This solution is based on the equation, $[H^{+}] = 10^{-pH}$ and because hydrochloric is a strong acid, it can be assumed to completely dissociate in solution.

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Example Question #112 : Gre Subject Test: Chemistry

What is the pH of a $2.0*10^{-4} M$ solution of NaOH ?

Possible Answers:

10.3

7.8

Correct answer:

10.3

Explanation:

We need to calculate the pH of a $2*10^{-4} M$ NaOH solution. There is one mole of OH^- in every mole of NaOH , therefore:

$[OH^{-}]=2*10^{-4}\ M$

$pOH=-log(OH^{-})=-log(2*10^{-4})=3.7$

The equation with the relationship between pH and pOH is:

pH+pOH=14

We can calculate the pH by rearranging this equation:

pH= 14.00-3.7=10.3

Another way of solving this problem is shown below. The equation with the relationship between $H_{3}O^{+}$ and $OH^{-}$ concentration is:

$K_{w}=[H_{3}O^{+}][OH^{-}]=1.0*10^{-14}$

Rearrange this equation:

$[H_{3}O^{+}]=\frac{[K_{w}]}{[OH^{-}]}=\frac{1.0*10^{-14}}{2*10^{-4}}=5*10^{-11}$

We can calculate the pH of this solution using the equation below:

$pH=-log(H_{3}O^{+})=-log(5.0*10^{-11})=10.3$

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Example Question #113 : Gre Subject Test: Chemistry

What is the pH of a $5.0*10^{-4} M$ solution of $Ca(OH)_{2}$ ?

Possible Answers:

Correct answer:

Explanation:

First we need to calculate the [OH^-] of the $5*10^{-4} M$ $Ca(OH)_{2}$ solution. For every mole of $Ca(OH)_{2}$ , there is double the number of moles of hydroxide ions:

$[OH^{-}]=2*(5 *10^{-4} M)=0.001M$

The pOH can be calculated using the below equation:

$pOH=-log(OH^{-})=-log(0.001)=3$

The equation with the relationship between pH and pOH is below:

pH+pOH=14

We can calculate the pH by rearranging this equation:

pH= 14.00-3.00=11

Another way of solving this problem is shown below. The equation with the relationship between $H_{3}O^{+}$ and $OH^{-}$ concentration is:

$K_{w}=[H_{3}O^{+}][OH^{-}]=1.0*10^{-14}$

Rearrange this equation:

$[H_{3}O^{+}]=\frac{[K_{w}]}{[OH^{-}]}=\frac{1.0*10^{-14}}{0.001}=1*10^{-11}$

We can calculate the pH of this solution using the equation below:

$pH=-log(H_{3}O^{+})=-log(1.0*10^{-11})=11$

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Example Question #2 : P H

Considering the K_a of (hydrofluoric acid) is $6.8\times10^{-4}$ , what is the pK_b of the base F^- ?

Possible Answers:

8.4

6.4

10.8

5.5

7.5

Correct answer:

10.8

Explanation:

The relationship between K_a and K_b is:

$K_{a}*\ K_{b}=K_{w}$

$K_{w}=1.0\times10^{-14}$

Rearranging this equation gives:

$K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0\times 10^{-14}}{6.8\times 10^{-4}}=1.5\times10^{-11}$

In order to calculate the pK_b , we must use this relationship:

$pK_{b}=-log(K_{b})=-log(1.5\times 10^{-11})=10.8$

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Example Question #31 : Acid Base Chemistry

What is the pH of a $0.05 \textup{ M}$ solution of HCl ?

Possible Answers:

11.0

0.05

5.5

4.1

1.3

Correct answer:

1.3

Explanation:

Below is the equilibria of HCl in an aqueous solution:

$HCl(l) +H_{2}O(l)\rightleftharpoons\ Cl^{-}(aq)+H_{3}O^{+}(aq)$

HCl is a strong acid so it completely ionizes in solution. It has a high $K_{eq}$ of ^.

There is 100% dissociation of HCl , therefore the resulting $H_{3}O^+$ in solution equals to the concentration of original HCl .

$[H_{3}O^{+}]=0.05M$

$pH=-log[H_{3}O^{+}]=-log[0.05]=1.3$

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Example Question #31 : Acid Base Chemistry

What is the pH of a $0.10 \textup{ M}$ solution of NaOH ?

Possible Answers:

5.0

8.1

1.0

0.1

Correct answer:

Explanation:

Below is the equilibria of NaOH in an aqueous solution:

$NaOH(s)\rightleftharpoons\ Na^{+}(aq)+OH^{-}(aq)$

NaOH is a strong base so it completely ionizes in solution. It has a high solubility product constant.

There is 100% dissociation of NaOH , therefore the resulting $OH^{-}$ in solution equals to the concentration of original NaOH .

$[OH^{-}]=0.10M$

$pOH=-log[OH^{-}]=-log[0.10]=1$

pH+pOH=14

pH=14-pOH=14-1=13

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Example Question #1 : P H

If $3.11\times10^{-4}\ moles$ of NaOH is reacted with $5.71\times10^{-4}\ moles$ of HCl , what is the pH if the total volume of the solution is 1 liter ?

Possible Answers:

2.1

3.6

8.4

6.6

Correct answer:

3.6

Explanation:

Based on the chemical equation, HCl and NaOH react in a 1:1 mole ratio. Therefore, in order to find the pH of this solution you must first determine the difference in moles of the two reactants and the limiting reactant. The reactant in excess will determine the pH of the solution. $5.71\times 10^{-4}\ moles\ HCl-3.11\times 10^{-4}\ moles\ NaOH=2.59\times 10^{-4}\ moles\ HCl$

There HCl is in excess and the pH of the solution will be based on the moles of the excess HCl .

$moles\ of\ H^{+}=2.59\times 10^{-4}}\ moles$

$Molarity\ [H^+]=\frac{moles\ of\ solute}{Liters\ of\ solution}=\frac{2.6\ moles}{1L}=2.6M$

$pH=-log[H^{+}]=-log[2.59\times 10^{-4}]=3.6$

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GRE Subject Test: Chemistry : General Chemistry

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #3 : Titrations

Example Question #112 : General Chemistry

Example Question #11 : Titrations

Example Question #21 : Acid Base Chemistry

Example Question #112 : Gre Subject Test: Chemistry

Example Question #113 : Gre Subject Test: Chemistry

Example Question #2 : P H

Example Question #31 : Acid Base Chemistry

Example Question #31 : Acid Base Chemistry

Example Question #1 : P H

All GRE Subject Test: Chemistry Resources

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