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GRE Subject Test: Chemistry : Acid-Base Reactions

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Example Question #11 : Acid Base Reactions

$H_{2}SO_{4}_{(aq)}+2NaOH_{(aq)}\rightarrow\ 2H_{2}O_{(l)}+Na_{2}SO_{4}_{(aq)}$

Based on the above balanced equation for a neutralization reaction, what is the NaOH concentration of a NaOH solution if 25.0 mL of 0.100 M $H_{2}SO_{4}$ is needed to neutralize a 15 mL solution of NaOH ?

Possible Answers:

0.112M

0.333M

0.667M

2.25M

4.10M

Correct answer:

0.333M

Explanation:

We need to convert milliliters to liters:

$25.0\ mL \times\frac{1L}{1000mL}=0.0250\ L$

We need to determine the moles of $H_{2}SO_{4}$ using dimensional analysis the concentration as a conversion factor:

$Moles\ of\ H_{2}SO_{4}=0.0250\ L \times \frac{0.100\ moles}{L}=0.00250\ moles\ H_{2}SO_{4}$

Based on the chemical equation given:

$1\ mole\ of H_{2}SO_{4}=2\ moles\ of\ NaOH$

We can used the relationship of moles of NaOH and moles of $H_{2}SO_{4}$ as a conversion factor to determine the moles of NaOH :

$Moles\ of\ NaOH=0.00250\ moles\ H_{2}SO_{4} \times\frac{2\ moles\ NaOH}{1\ mole\ H_{2}SO_{4}}=0.00500\ moles\ NaOH$

Concentration in molarity can be calculated using the following formula:

$Molarity\ of NaOH=\frac{moles\ of\ NaOH}{Liters\ of\ solution}$

Let's convert liters of NaOH to the mL:

$Liters\ of\ NaOH\ solution=15.0\ mL \times\frac{1L}{1000mL}=0.0150\ L$

Therefore, the concentration of NaOH is:

$Molarity\ of NaOH=\frac{0.00500\ moles\ NaOH}{0.0150\ L}=0.333\ M$

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Example Question #12 : Acid Base Reactions

$HNO_{3}_{(aq)}+NaOH_{(aq)}\rightarrow\ H_{2}O_{(l)}+NaNO_{3}_{(aq)}$

Based on the above balanced equation for a neutralization reaction, how many moles of $NaNO_{3}$ is formed if 20.0 mL of 0.100 M $HNO_{3}$ is needed to neutralize a solution of NaOH ?

Possible Answers:

0.120mol

0.00200mol

0.0450mol

0.200mol

0.00400mol

Correct answer:

0.00200mol

Explanation:

We need to convert milliliters of $HNO_{3}$ to liters:

$20.0\ mL \times\frac{1L}{1000mL}=0.0200\ L$

We need to determine the moles of $HNO_{3}$ using dimensional analysis and the concentration as a conversion factor:

$Moles\ of\ HNO_{3}=0.0200\ L \times \frac{0.100\ moles}{L}=0.00200\ moles\ HNO_{3}$

Based on the chemical equation given:

$1\ mole\ of HNO_{3}=1\ mole\ of\ NaNO_{3}$

We can use the relationship of moles of $HNO_{3}$ and moles of $NaNO_{3}$ as a conversion factor to determine the moles of $NaNO_{3}$ :

$Moles\ of\ NaNO_{3}=0.00200\ moles\ HNO_{3} \times\frac{1\ moles\ NaNO_{3}}{1\ mole\ HNO_{3}}=0.00200\ moles\ NaNO_{3}$

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Example Question #11 : Acid Base Chemistry

If 4.2 moles of NaOH is reacted with 3.5 moles of HCl , how many grams of NaCl is produced?

Possible Answers:

69.3g

204.5g

23.1g

10.2g

Correct answer:

204.5g

Explanation:

HCl is the limiting reactant because based on the equation, HCl and NaOH react in a 1:1 mole ratio. However, there is 0.5 moles more of NaOH than HCl therefore the amount of NaCl produced is limited by the amount of HCl present in the reaction. Based on the equation for every 1 mole of HCl reacted, 1 mole of NaCl was produced:

$moles\ of\ HCl\ reacted=moles\ of\ NaCl\ produced$

Therefore,

$3.5\ moles\ NaCl\times \frac{58.44g}{moles}=204.5g\ NaCl$

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Example Question #101 : General Chemistry

Determine the acid dissociation constant for a $0.1\ M$ monoprotic acid $\left ( HA\right )$ dissolved in the $H_{2}O$ in which a pH meter for read pH 3.3 for the solution.

Possible Answers:

$1.52\times10^{-3}$

$2.51\times10^{-6}$

$3.3\times10^{-2}$

$7.9\times10^{-12}$

Correct answer:

$2.51\times10^{-6}$

Explanation:

The acid dissociation constant expression for this reaction is: $K_{a}=\frac{[H_{3}O^{+}][A^{-}]}{[HA]}$

Due to the dissociation of :

$x=[H_{3}O^{+}]=[A^{-}]$

Using the following equation, we can calculate the $H_{3}O^{+}$ concentration:

$[H_{3}O^{+}]=10^{-pH}$

$[H_{3}O^{+}]=10^{-3.3}$

$[H_{3}O^{+}]=5.01\times10^{-4}M$

$[HA]=original\ HA\ concentration-x=0.1M-x\approx0.1M$

Plug the values obtained in to the acid dissociation constant expression:

$K_{a}=\frac{[H_{3}O^{+}][A^{-}]}{[HA]}=\frac{x\times x}{0.1\ M}=\frac{(5.01\times10^{-4})^{2}M}{0.1\ M}=2.51\times10^{-6}$

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GRE Subject Test: Chemistry : Acid-Base Reactions

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

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #11 : Acid Base Reactions

Example Question #12 : Acid Base Reactions

Example Question #11 : Acid Base Chemistry

Example Question #101 : General Chemistry

All GRE Subject Test: Chemistry Resources

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