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College Chemistry : Reactions

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Example Question #1 : Solubility Product Constant

How many grams of $\textrm{BaF}_{\textrm{2}}$ are dissolved in 75.0 mL of a saturated solution of $\textrm{BaF}_{\textrm{2}}$ ?

$\textrm{BaF}_{\textrm{2}}(s)\rightleftharpoons\textrm{Ba}^{\textrm{2+}}(aq)\textrm{ + 2 F}^{\textrm{ -}}(aq)$

$\textrm{K}_{\textrm{sp}}=1.00\textrm{ x 10}^{-6}$

Possible Answers:

$0.0132\textrm{ g BaF}_{2}$

$0.0829\textrm{ g BaF}_{2}$

$0.166\textrm{ g BaF}_{2}$

$0.132\textrm{ g BaF}_{2}$

$1.10\textrm{ g BaF}_{2}$

Correct answer:

$0.0829\textrm{ g BaF}_{2}$

Explanation:

To calculate how many grams of $\textrm{BaF}_{\textrm{2}}$ are dissolved in the solution, we first solve for the molarity of the solution.

Using the dissociation equation

$\textrm{BaF}_{\textrm{2}}(s)\rightleftharpoons\textrm{Ba}^{\textrm{2+}}(aq)\textrm{ + 2 F}^{\textrm{ -}}(aq)$

We can write out the equation for the solubility product constant as

$\textrm{K}_{\textrm{sp}}=\textrm{[Ba}^{2+}][\textrm{F}^{\textrm{ -}}]^{2}$

$1.00\textrm{ x 10}^{-6}=\textrm{[Ba}^{2+}][\textrm{F}^{\textrm{ -}}]^{2}$

Because there are 2 fluoride ions for every bariumion, we can rewrite the equation as

$[\textrm{Ba}^{2+}]=x$

$1.00\textrm{ x 10}^{-6}=(x)(2x)^{2}$

Now solve for x

$1.00\textrm{ x 10}^{-6}=(x)(4x^{2})$

$1.00\textrm{ x 10}^{-6}=4x^{3}$

$x^{3}=\frac{1.00\textrm{ x 10}^{-6}}{4}=2.50\textrm{ x 10}^{-7}$

$x=\sqrt[3]{2.50\textrm{ x 10}^{-7}}=6.30\textrm{ x 10}^{-3}$

$[\textrm{Ba}^{2+}]=6.30\textrm{ x 10}^{-3}M$

Solve for the concentration of dissolved $\textrm{BaF}_{\textrm{2}}$

$[\textrm{BaF}_{2}]=[\textrm{Ba}^{2+}]=6.30\textrm{ x 10}^{-3}M$

Now that we have the concentration of dissolved $\textrm{BaF}_{\textrm{2}}$ , we can calculate how many grams are dissolved in the solution

$(6.30\textrm{ x 10}^{-3}M)(0.0750\textrm{ L})=4.73\textrm{ x 10}^{-4}\textrm{ mol}$

$(4.73\textrm{ x 10}^{-4}\textrm{ mol})(175.34\frac{\textrm{g}}{\textrm{mol}})=0.0829\textrm{ g BaF}_{2}$

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Example Question #141 : College Chemistry

Which of the following acid and base pairs are capable of acting as a buffer?

Possible Answers:

HCl and $Cl^{-}$

$HNO_{3}$ and $NO_{3}^{-}$

$H_{2}SO_{4}$ and $HSO_{4}^{-}$

$H_{2}CO_{3}$ and $HCO_{3}^{-}$

Correct answer:

$H_{2}CO_{3}$ and $HCO_{3}^{-}$

Explanation:

In this question, we're presented with a variety of acid/base pairs and we're asked to identify which one could act as a buffer.

Remember that a buffer is a pair of acid and its conjugate base that acts to resist substantial changes in pH. In order for a buffer to work, the acid base pair needs to exist in equilibrium. This way, when the pH of the solution changes, the equilibrium of the acid/base reaction will shift, such that the pH will not change drastically.

To have an acid/base pair in equilibrium, we'll need to look for a pair that contains a weak acid. Acids like HCl and $H_{2}SO_{4}$ are so strong that they will dissociate completely. Of the answer choices shown, only the carbonic acid/bicarbonate system ( $H_{2}CO_{3}$ and $HCO_{3}^{-}$ ) exists in equilibrium. Thus, this is the correct answer.

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Example Question #1 : Acid Base Reactions

You are in chemistry lab performing a titration. You were given 15 mL of an aqueous solution with an unknown concentration of acetic acid, CH_3COOH to solve through titration with concentrated sodium hydroxide, NaOH . You know that the pKa of acetic acid is 4.75 and that your titrant is 0.1 M sodium hydroxide, NaOH .

The endpoint was determined at 10 mL of sodium hydroxide, NaOH . What is the pH after 5 mL of NaOH was added?

Possible Answers:

8.45

9.25

4.75

Correct answer:

4.75

Explanation:

At the half end point, the pH=pKa . This can be determined by the Henderson-Hasselbalch equation if it is not clear.

Since the endpoint of the titration is that there are 10 mL of 0.1 M NaOH added, that means that there are 0.001 moles of acetic acid.

When 5 mL of NaOH is added, there are 0.0005 moles of acetic acid and 0.0005 moles of acetate formed.

$pH = pKa + log \frac{base}{acid} = pKa + log \frac{0.0005}{0.0005 }$

log(1) = 1

Therefore pH= pKa

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Example Question #51 : Reactions

Determine which combination of solutions would create a buffer solution.

Possible Answers:

25 mL of 0.4 M NaOH + 50 mL of 0.2 M ammonium

10 mL of 0.4 M NaOH + 10 mL of 0.2 M ammonium

50 mL of 0.4 M NaOH + 50 mL of 0.8 M ammonium

5 mL of 0.4 M NaOH + 10 mL of 0.2 M ammonium

Correct answer:

50 mL of 0.4 M NaOH + 50 mL of 0.8 M ammonium

Explanation:

For all the other options there is no ammonium leftover with which to serve as the weak acid in the buffer system, the ammonium is all used up and converted to ammonia. However in the correct answer choice, there is enough ammonium leftover after the reaction with the sodium hydroxide.

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Example Question #1 : Buffers

Determine the pH of an aqueous solution of 0.01 M acetic acid, $CH_{}3COOH$ . The pKa of acetic acid is 4.75.

Possible Answers:

2.87

3.45

0.161

Correct answer:

2.87

Explanation:

Since acetic acid is a weak acid, it has a Ka that is rather small, we have to do a RICE table to determine the equilibrium amount of hydronium, H₃O⁺ to then determine the pH.

R $CH_{}3COOH (aq) + H_{}2O (l) \rightarrow CH_{}3COO^{}- (aq) + H_{}3O^{}+ (aq)$

I 0.1 M - 0 0

C -x +x +x

E 0.1 -x x x

So first we need to change our pKa to a Ka

where $Ka = 10^{}-^{^{^{}}}pKa$ therefore

$Ka= 10^{}-^{}4^{}.^{}7^{}5 = 1.78 x 10^{}-^{}5$

= $\frac{[CH_{}3COO^{}-][H_{}3O^{}+]}{[CH_{}3COOH]}$ = $\frac{x*x}{(0.1-x)}$

If we assume that x is very small compared to 0.1...

$1.78x10^{}-5= \frac{x*x}{0.1}$

Where x = 0.00133

(note: when solving using the quadratic we come up with the same answer)

So if $x= 0.00133 = [H_{}3O^{}+]$

$pH = -log [H_{}3O^{}+] =-log(0.0013) = 2.87$

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Example Question #52 : Reactions

Determine which solution(s) will yield a buffer solution.

I. 10 mL of 0.5 M HCl + 20 mL of 0.5 M acetate

II. 10 mL of 0.5 M HCl + 10 mL of 0.5 M acetate

III. 10 mL of 0.5 M HCl + 10 mL of 1.0 M acetate

IV. 10 mL of 0.5 M HCl + 10 mL of 1.5 M acetate

Possible Answers:

I, III, and IV

I, II, III, and IV

I only

III and IV

I and III

Correct answer:

I, III, and IV

Explanation:

These answers are correct because the two components needed to create a buffer solution are a weak acid and its conjugate base, or a weak base and its conjugate acid. In these cases, the first reaction to occur upon addition of the strong acid is the formation of the conjugate acid, acetic acid.

$HCl + CH_3COO^- \rightarrow CH_3COOH$

If the amount of initial CH_3COO^- is greater than HCl, then we will have some left over to act as a buffer with the created conjugate acid. This can be through a greater volume, or through a higher concentration as shown in the correct answers.

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Example Question #53 : Reactions

Which combination(s) would create a buffer solution?

I. Weak acid

II. Weak acid's conjugate base

III. Strong acid

IV. Strong base

V. Weak base

VI. Weak base's conjugate acid

Possible Answers:

I and II; III and IV; V and VI

None of these will combine to form a buffer solution

I and II; V and VI

III and IV only

Correct answer:

I and II; V and VI

Explanation:

A buffer solution is formed from the equilibrium of a weak acid and its conjugate base, or from a weak base and its conjugate acid. It's ability to "buffer" the pH or keep it from changing in large amounts in from the switching between these two forms weak and its conjugate.

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Example Question #54 : Reactions

Determine which of these solution combinations form a buffer.

Possible Answers:

10 mL of 0.5 M NaOH + 20 mL of 0.5 M ammonium

10 mL of 0.5 M NaOH + 10 mL of 0.5 M ammonium

10 mL of 0.5 M NaOH + 10 mL of 0.5 M HCl

10 mL of 0.5 M NaOH + 10 mL of 0.5 M ammonia

Correct answer:

10 mL of 0.5 M NaOH + 20 mL of 0.5 M ammonium

Explanation:

First to go through why the other ones are wrong:

Strong base + strong acid neutralizes and does not form a buffer solution

Strong base + weak base does not form a buffer - would need an acid

Strong base + weak acid = all weak acid converted to conjugate base

The correct answer is:

Strong base + weak acid = half converted to conjugate base with half leftover as weak acid, with all the components for a buffer

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Example Question #1 : Titrations

The titration of a $15.0\text{mL}$ sample of an $\text{HCl}$ solution at an unknown concentration requires $17.08\text{mL}$ of a $0.200\text{M NaOH}$ solution. What is the concentration of the unknown hydrochloric acid solution in ?

Possible Answers:

0.245

0.189

0.237

0.228

Correct answer:

0.228

Explanation:

Start by writing the chemical equation for this acid-base reaction.

$\text{HCl}(aq)+\text{NaOH}(aq)\rightarrow\text{H}_2\text{O}(l)+\text{NaCl}(aq)$

Next, find the number of moles of $\text{NaOH}$ that was added.

$17.08\text{mL NaOH}\cdot\frac{1\text{L}}{1000\text{mL}}\cdot\frac{0.200\text{ moles NaOH}}{\text{L}}=0.003416\text{ moles of NaOH}$

At the equivalence point of a titration, the moles of acid and the moles of base are the same.

$\text{Moles of HCl}=\text{Moles of NaOH}=0.003416$

Finally, find the molarity of the solution.

$\text{Molarity}=\frac{\text{moles of solute}}{\text{liters of solution}}=\frac{0.003416}{15.0\times10^{-3}}=0.228M$

Make sure to round the answer to three significant figures.

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Example Question #1 : Titrations

A $75.0\text{mL}$ sample of 0.100M sodium hydroxide is titrated with 0.150M hydrochloric acid. Find the pH of the solution after adding $45.0\text{mL}$ of hydrochloric acid.

Possible Answers:

12.2

10.4

11.8

11.2

Correct answer:

11.8

Explanation:

Start by finding the initial number of moles of sodium hydroxide by using the given volume and molarity. Since sodium hydroxide is a strong base, it will dissociate completely into $\text{OH}^-$ .

$\text{Initial moles of NaOH}=\text{Initial moles of OH}^-=0.075\text{L}(\frac{0.100\text{ mol}}{\text{L}})=0.0075\text{ moles}$

Next, find the number of moles of the hydrochloric acid added by using the given volume and molarity. Since hydrochloric acid is a strong acid, all of the $\text{H}_3\text{O}^+$ will come from the acid.

$\text{Moles of HCl added}=\text{Moles of H}_3\text{O}^+=0.045\text{L}(\frac{0.150\text{ moles}}{\text{L}})=0.00675\text{ moles}$

Now, as the acid is added, it will neutralize some of the base as shown by the following equation:

$\text{OH}^-(aq)+\text{H}_3\text{O}^+(aq)\rightarrow 2\text{H}_2\text{O}(l)$

Since the reactants are found in a 1:1 ratio, subtract the number of moles of acid added from the number of initial moles of base to get the amount of remaining base.

$\text{Remaining moles of NaOH}=0.0075-0.00675=7.5\times10^{-4}\text{ moles of NaOH}$

Now, divide this number of moles by the total volume of the solution to find the concentration of the base.

$[\text{OH}]^-=\frac{7.5\times 10^{-4}\text{ moles}}{0.075\text{L}+0.045\text{L}}=0.00625\text{ M}$

Next, find the $\text{pOH}$ value.

$\text{pOH}=-\log(0.00625)=2.204$

Finally, find the $\text{pH}$ value.

$\text{pH}=14-\text{pOH}$

$\text{pH}=14-2.204=11.80$

Your answer should have significant figures.

$\text{pH}=11.8$

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College Chemistry : Reactions

Study concepts, example questions & explanations for College Chemistry

All College Chemistry Resources

Example Questions

Example Question #1 : Solubility Product Constant

Example Question #141 : College Chemistry

Example Question #1 : Acid Base Reactions

Example Question #51 : Reactions

Example Question #1 : Buffers

Example Question #52 : Reactions

Example Question #53 : Reactions

Example Question #54 : Reactions

Example Question #1 : Titrations

Example Question #1 : Titrations

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