A 50.0 mL solution of 0.129 M KOH is titrated with 0.258 M HCl. Calculate the pH of the solution after the addition of each of the given amounts of HCl.
A) 0.00ml
B)7.00ml
C)12.5ml
D)18.0ml
E)24.0ml
F)25.0ml
G)26.0ml
H)29.0ml

please show work with steps .

Answers

Answer 1

Answer:

A- pH = 13.12

B- pH = 12.91

C- pH = 12.71

D- pH = 12.43

E- pH = 11.55

F- pH = 7

G- pH = 2.46

H- pH = 1.88

Explanation:

This is a titration of a strong base with a strong acid. The neutralization reaction is: KOH (aq) + HCl (aq) →  H₂O(l) + KCl(aq)

Our pH at the equivalence point is 7, because we have made a neutral salt.

To determine the volume at that point we state the formula for titration:

mmoles of base = mmoles of acid

Volume of base  . M of base = Volume of acid . M of acid

50mL . 0.129M = 0.258 M . Volume of acid

Volume of acid = (50mL . 0.129M) / 0.258 M →  25 mL (Point F)

When we add 25 mL of HCl, our pH will be 7.

A- At 0 mL of acid, we only have base.

KOH → K⁺ + OH⁻

[OH⁻] = 0.129 M

To make more easy the operations we will use, mmol.

mol . 1000 = mmoles → mmoles / mL = M

- log 0.129 = 0.889

14 - 0.889 = 13.12

B-  In this case we are adding, (7 mL . 0.258M) = 1.81 mmoles of H⁺

Initially we have  0.129 M . 50 mL = 6.45 mmoles of OH⁻

1.81 mmoles of H⁺ will neutralize, the 6.45 mmoles of OH⁻ so:

6.45 mmol - 1.81 = 4.64 mmoles of OH⁻

This mmoles of OH⁻ are not at 50 mL anymore, because our volume has changed. (Now, we have 50 mL of base + 7 mL of acid) = 57 mL of total volume.

[OH⁻] = 4.64 mmoles / 57 mL = 0.0815 M

- log 0.0815 M = 1.09 → pOH

pH = 14 - pOH → 14 - 1.09 = 12.91

C- In this case we add (12.5 mL . 0.258M) = 3.22 mmoles of H⁺

Our initial mmoles of OH⁻ would not change through all the titration.

Then 6.45 mmoles of OH⁻ are neutralized by 3.22 mmoles of H⁺.

6.45 mmoles of OH⁻ - 3.22 mmoles of H⁺ = 3.23 mmoles of OH⁻

Total volume is: 50 mL of base + 12.5 mL = 62.5 mL

[OH⁻] = 3.23 mmol / 62.5 mL = 0.0517 M

- log  0.0517 = 1.29 → pOH

14 - 1.11 = 12.71

D- We add (18 mL . 0.258M) = 4.64 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 4.64 mmoles of H⁺.

6.45 mmoles of OH⁻ - 4.64 mmoles of H⁺ = 1.81 mmoles of OH⁻

Total volume is: 50 mL of base + 18 mL = 68 mL

[OH⁻] = 1.81 mmol / 68 mL = 0.0265 M

- log  0.0265 = 1.57 → pOH

14 - 1.57 = 12.43

E- We add (24 mL . 0.258M) = 6.19 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 6.19 mmoles of H⁺.

6.45 mmoles of OH⁻ - 6.19 mmoles of H⁺ = 0.26 mmoles of OH⁻

Total volume is: 50 mL of base + 24 mL = 74 mL

[OH⁻] = 0.26 mmol / 74 mL = 3.51×10⁻³ M

- log  3.51×10⁻³  = 2.45 → pOH

14 - 2.45 = 11.55

F- This the equivalence point.

mmoles of OH⁻ = mmoles of H⁺

We add (25 mL . 0.258M) = 6.45 mmoles of H⁺

All the OH⁻ are neutralized.

OH⁻  +  H⁺  ⇄   H₂O              Kw

[OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷  →  pOH = 7

pH → 14 - 7 = 7

G- In this case we have an excess of H⁻

We add (26 mL . 0.258M ) = 6.71 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

6.71 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 0.26 mmoles of H⁺

[H⁺] = 0.26 mmol / Total volume

Total volume is: 50 mL + 26 mL → 76 mL

[H⁺] = 0.26 mmol / 76 mL → 3.42×10⁻³ M

- log 3.42×10⁻³ = 2.46 → pH

H- Now we add (29 mL . 0.258M) = 7.48 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

7.48 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 1.03 mmoles of protons

Total volume is 50 mL + 29 mL = 79 mL

[H⁺] = 1.03 mmol / 79 mL → 0.0130 M

- log 0.0130 = 1.88 → pH

After equivalence point, pH will be totally acid, because we always have an excess of protons. Before the equivalence point, pH is basic, because we still have OH⁻ and these hydroxides, will be neutralized through the titration, as we add acid.

Answer 2

The pH value is given by 14 less the logarithm of the hydronium ion

concentration or the hydrogen ion concentration in the solution.

Responses (approximate values):

The pH values are;

A) 13.11

B) 12.91

C) 12.71

D) 12.42

E) 11.54

F) 7

G) 2.469

H) 1.884

Which is used to find the pH of the solution?

A) Concentration of the KOH = 0.129 M

Amount of HCl added = 0.00 ml

The pH = -log[H⁺] = 14 - pOH

pOH = -log[OH⁻]

Which gives;

pH = 14 - (-log[OH⁻] )

pH = 14 - (-log(0.129)) ≈ 13.11

B) Volume of acid added = 7.00 mL = 0.007 L

Concentration of the acid = 0.258 M HCl

Number of moles of acid, H⁺ = 0.007 × 0.258 moles = 0.001806 moles

Number of moles of KOH remaining, OH⁻= 0.05 × 0.129 - 0.001806 = 0.004644

Number of moles of OH⁻ = 0.004644 moles

[tex]Concentration, \ [OH^-] = \dfrac{0.004644 \, moles}{0.05 7 \, L} \approx \mathbf{ 0.0815 \, M}[/tex]

pH of solution = 14 - (-log(0.0815)) ≈ 12.91

C) 12.5 mL HCl contains, 0.0125 × 0.258 moles = 0.003225 moles

OH⁻ remaining = 0.05 × 0.129 - 0.003225 = 0.003225 moles

[tex]Concentration \ of \ [OH^-]= \dfrac{0.003225\, moles}{(0.05 + 0.0125) \, L} = \mathbf{0.0516 \, M}[/tex]

pH of solution = 14 - (-log(0.0516)) ≈ 12.71

D) 18.0 mL HCl contains, 0.018 × 0.258 moles = 0.004644 moles

OH⁻ remaining = 0.05 × 0.129 - 0.004644 = 0.001806 moles

[tex]Concentration \ of \ [OH^-] = \dfrac{0.001806\, moles}{(0.05 + 0.018) \, L} \approx \mathbf{0.0266\, M}[/tex]

pH of solution = 14 - (-log(0.0266)) ≈ 12.42

E) 24.0 mL HCl contains, 0.024 × 0.258 moles = 0.006192 moles

OH⁻ ion remaining = 0.05 × 0.129 - 0.006192 = 0.000258 moles

[tex]Concentration \ of \ [OH^-] = \dfrac{0.000258\, moles}{(0.05 + 0.024) \, L} \approx \mathbf{0.0035\, M}[/tex]

pH of solution = 14 - (-log(0.0035)) ≈ 11.54

F) 25.0 mL HCl contains, 0.025 × 0.258 moles = 0.00645 moles

[OH⁻] remaining = 0.05 × 0.129 - 0.00645 = 0 moles

[tex]Concentration \ of \ [OH^-] = \dfrac{0\, moles}{(0.05 + 0.024) \, L} \approx 0\, M[/tex]

Therefore;

Number of moles of KOH = 0, or the solution is neutralized

[OH⁻] = [H⁺]

Which gives;

pH = pOH = 7

G) 26.0 mL HCl contains, 0.026 × 0.258 moles = 0.006708 moles

[OH⁻] remaining = 0.05 × 0.129 - 0.006708 = -0.000258 moles

Therefore

Number of moles of H⁺ = 0.000258

[tex]\mathbf{Concentration} \ of \ \mathbf{[H^+] }= \dfrac{0.000258\, moles}{(0.05 + 0.026) \, L} \approx 0.003395\, M[/tex]

pH of solution = (-log(0.003395)) ≈ 2.469

H) 29.0 mL HCl contains, 0.029 × 0.258 moles = 0.007482 moles

H⁺ remaining = 0.007482 - 0.05 × 0.129 = 0.001032 moles

Therefore

Number of moles of H⁺ = 0.001032

[tex]Concentration \ of \ [H^+] = \mathbf{ \dfrac{0.001032\, moles}{(0.05 + 0.029) \, L}} \approx 0.01306\, M[/tex]

pH of solution = (-log(0.01306)) ≈ 1.884

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Answer:

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Reaction Mole RatiosUsing Dimensional AnalysisExplanation:

Step 1: Define

[RxN - Unbalanced] Mg (s) + HCl (aq) → MgCl (aq) + H₂ (g)

[RxN - Balanced] 2Mg (s) + 2HCl (aq) → 2MgCl (aq) + H₂ (g)

[Given] 3.25 mol Mg

[Solve] x g HCl

Step 2: Identify Conversions

[RxN] 2 mol Mg → 2 mol HCl

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of HCl - 1.01 + 35.45 = 36.46 g/mol

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[S - DA] Set up:                                                                                                 [tex]\displaystyle 3.25 \ mol \ Mg(\frac{2 \ mol \ HCl}{2 \ mol \ Mg})(\frac{36.46 \ g \ HCl}{1 \ mol \ HCl})[/tex][S - DA] Multiply/Divide [Cancel out units]:                                                    [tex]\displaystyle 127.61 \ g \ HCl[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

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Answers

Answer:

A. 0.0625 moldm¯³

B. 6.125 gdm¯³

Explanation:

We'll begin by writing the balanced equation balance equation for the reaction. This is illustrated below:

2NaOH + H₂SO₄ —> Na₂SO₄ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, H₂SO₄ (nₐ) = 1

The mole ratio of the base, NaOH (n₆) = 2

A. Determination of the concentration of the acid in moldm¯³.

Volume of base, NaOH (V₆) = 20 cm³

Concentration of base, NaOH (C₆) = 0.2 moldm¯³

Volume of acid, H₂SO₄ (Vₐ) = 32 cm³

The mole ratio of the acid, H₂SO₄ (nₐ) = 1

The mole ratio of the base, NaOH (n₆) = 2

Concentration of acid, H₂SO₄ (Cₐ) =?

CₐVₐ / C₆V₆ = nₐ / n₆

Cₐ × 32 / 0.2 × 20 = 1 / 2

Cₐ × 32 / 4 = 1 / 2

Cross multiply

Cₐ × 32 × 2 = 4 × 1

Cₐ × 64 = 4

Divide both side by 64

Cₐ = 4 / 64

Cₐ = 0.0625 moldm¯³

Therefore, concentration of the acid in moldm¯³ is 0.0625 moldm¯³

B. Determination of the concentration of the acid in gdm¯³.

Concentration (moldm¯³) = 0.0625 moldm¯³

Concentration (gdm¯³) =?

Next, we shall determine the molar mass of the acid, H₂SO₄. This can be obtained as follow:

Molar mass of H₂SO₄ = (2×1) + 32 + (4×16)

= 2 + 32 + 98

= 98 g/mol

Finally, we shall determine the concentration of the acid in gdm¯³. This can be obtained as follow:

Concentration (moldm¯³) = 0.0625 moldm¯³

Molar mass of H₂SO₄ = 98 g/mol

Concentration (gdm¯³) =?

Conc. (moldm¯³) = conc. (gdm¯³) / Molar mass

0.0625 = Conc. (gdm¯³) / 98

Cross multiply

Conc. (gdm¯³) = 0.0625 × 98

Conc. (gdm¯³) = 6.125 gdm¯³

Thus, the concentration of the acid in gdm¯³ is 6.125 gdm¯³

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Required:
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b. Does the temperature of the water bath go up or down?

Answers

Answer:

The reaction is exothermic

The temperature of the water bath goes up

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An exothermic reaction is one in which energy flows out of the reaction system.

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As energy is lost to the surroundings, the temperature of the water bath rises accordingly.

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Explanation:

Other Questions
Explain the process of protein synthesis be sure to include the following terms ribosome and mRNA TRNA codon and peptide bonds One leg of a right triangle is 6 units long, and its hypotenuse is 12 units long. What is the length of the other leg? Round to the nearest whole number. you are cinsidering investing in a peice of equipment to implement a cost cutting proposal the pre tax cost reduction is expected to equal $41.67 for each of the three years of the project's life. The equipment has an initial cost of $125 and belongs in a 20% CCA class. Assume a 34% tax bracket, a discount rate of 15%, and a salvage value of zero. If the equipment is sold to another company at the end of year 3 for $20, what is the PI? Which of the following is NOT a true statment regarding the Alaris MedSystem III? The table shows a probability distribution P(X) for a discrete random variable X. What is P(X>2)? Complete each statement by underlining the correct term or phrase in the brackets. A receptor is a [protein / fatty acid] to which a molecule binds. the tendency to see members of out-groups as very similar to each other is called: ___________________ homology includes amino acids and DNA sequencing, while homology includes features that are similar in their anatomy a locked section of fault is often identified by the existence of __________ there. TRUE OR FALSE you are more likely to find an item by using a binary search than by using a linear search. which of the following can be used to add visual elements with information that is part of the message? "jem condescended to take me to school the first day, a job usually done by ones parents...." according to the context, condescended means _____. Visual aids should serve a purpose beyond simply meeting audience desires. What is one of the purposes listed in the textbook?A.) To reinforce or emphasize a point.B.) To make the presentation more visually appealing.C.) To provide something for the audience to focus on when they get bored. Reread lines 49. Identify two examples of parallelism in these lines and describe the effect of each. order the following species found in the fossil record based on their age, starting with the oldest on top. Use the table of Consumer Price Index values and subway fares to determine a line of regression that predicts the fare when the CPI is given. CPI 30.2 48.3 112.3 162.2 191.9 197.8 Subway Fare 0.15 0.35 1.00 1.35 1.50 2.00 O j = 0.00955 0.124x O =-0.0331 +0.00254x O =-0.124 + 0.00955x O = 0.00254 0.0331x Using literature, describe how 31P NMR or other nuclei can be used for other quantitative measurements other than structure elucidation. Cite your source, which must be a primary resource. This is for Inorganic Chemistry Lab people demand freedom of speech as a compensation for the freedom of thought which they seldom use. tony and pepper are directors and shareholders of stark software, inc. tonys written authorization to pepper to vote tonys shares at a stark shareholders meeting is Create two lists. On one list, record the many ways that schools around 1900 are similar to the school you attend. On the otherecord the many differences.