The bond type in CaO is

Consider the titration of 40.0 mL of 0.200 M HClO4 by 0.100 M KOH. Calculate the pH of the resulting solution after the followin

olya-2409 [2.1K]

Answer:

a) 0.70

b) 7.00

c) 0.85

d) 12.15

e) 1.30

Explanation:

The neutralization reaction involved in the titration is:

HClO₄(aq) + KOH(aq) → KClO₄(aq) + H₂O(l)

According to the chemical equation, 1 mol of HClO₄ reacts with 1 mol of KOH (1 equivalent of acid with 1 equivalent of base). The moles are calculated from the product of the molar concentration (M) and the volume in liters.

We have the following moles of acid (HClO₄):

40.0 mL x 1 L/1000 mL = 0.04 L

0.200 mol/L x 0.04 L = 8 x 10⁻³ moles HClO₄

Since HClO₄ is a strong acid (completely dissociated into H⁺ and ClO₄⁻ ions), the moles of HClO₄ are equal to the moles of H⁺. Then, we can calculate the initial pH:

[H⁺] = 0.200 M → pH = -log [H⁺] = -log (0.200) = 0.70

Now, we calculate the pH after the addition of KOH. Since KOH is a strong base, the concentration of KOH is equal to the concentration of OH⁻ ions.

a) 0.0 mL

No KOH is added, so the pH is the initial pH: 0.70

b) 80.0 mL KOH

80.0 mL x 1 L/1000 mL = 0.08 L

0.100 mol/L x 0.08 L = 8 x 10⁻³ moles KOH = 8 x 10⁻³ moles OH⁻

After neutralization we have:

8 x 10⁻³ moles H⁺ - 8 x 10⁻³ moles OH⁻ = 0

The neutralization reaction is complete and there is no remaining H⁺ from the acid. The concentration of H⁺ is equal to the concentration of H⁺ of water:

[H⁺] = 1 x 10⁻⁷ M → pH = -log [H⁺] = -log (1 x 10⁻⁷) = 7.0

c) 10.0 mL KOH

10.0 mL x 1 L/1000 mL = 0.01 L

0.100 mol/L x 0.01 L = 1 x 10⁻³ moles KOH = 1 x 10⁻³ moles OH⁻

After neutralization we have:

8 x 10⁻³ moles H⁺ - 1 x 10⁻³ moles OH⁻ = 7 x 10⁻³ moles H⁺

The total volume is: V = 40.0 mL + 10.0 mL = 50 mL = 0.05 L

[H⁺] = 7 x 10⁻³ moles/0.05 L = 0.14 → pH = -log [H⁺] = -log (0.14) = 0.85

d) 100.0 mL KOH

100.0 mL x 1 L/1000 mL = 0.1 L

0.100 mol/L x 0.1 L = 0.01 moles KOH = 1 x 10⁻² moles OH⁻

After neutralization we have:

1 x 10⁻² moles OH⁻ - 8 x 10⁻³ moles H⁺ = 2 x 10⁻³ moles OH⁻

The total volume is: V = 40.0 mL + 100.0 mL = 140 mL = 0.14 L

[OH⁻] = 2 x 10⁻³ moles/0.14 L = 0.014 → pOH = -log [OH⁻] = -log (0.014) = 1.84

pH + pOH = 14 → pH = 14 - pOH = 14 - 1.84 = 12.15

e) 40.0 mL KOH

40.0 mL x 1 L/1000 mL = 0.04 L

0.100 mol/L x 0.04 L = 4 x 10⁻³ moles KOH = 4 x 10⁻³ moles OH⁻

After neutralization we have:

8 x 10⁻³ moles H⁺ - 4 x 10⁻³ moles OH⁻ = 4 x 10⁻³ moles H⁺

The total volume is: V = 40.0 mL + 40.0 mL = 80.0 mL = 0.08 L

[OH⁻] = 4 x 10⁻³ moles/0.08 L = 0.05 M → pH = -log [H⁺] = -log (0.05) = 1.30

5 0

3 years ago

What product would you obtain if you evaporate the water from the naoh layer prior to acidifying the layer?

AlekseyPX

The product that you obtain if you evaporate the water from the NaoH layer prior to acidifying the layer is salt of para - tery-butyphenol

Explanation

Para-tery-butyphenol can be isolate after extracting it into NaOH solution through carrying out

heating and then cooling,
then 3M of HCl is added to acidify.
Then ice bath is done if needed
finally the precipitate is filtered.

Para tery butyphenol can be used for production of alkylphenolic.

3 0

3 years ago

Read 2 more answers

Calculate the number atoms in a 92.8 mg of Si

Step2247 [10]

Answer: the answer is that stop using brainly for your answer dumb kid

Explanation:

dwcwcwcwcwce

8 0

4 years ago

What is the pH of an acetic acid solution where the concentration of acetic acid is 2 mM and the concentration of sodium acetate

schepotkina [342]

Answer:

pH = 5.76

Explanation:

We can solve this problem by using Henderson-Hasselbach's equation:

pH = pKa + log $\frac{[SodiumAcetate]}{[AceticAcid]}$

We are already know all the required information, thus we input the data given by the problem:

pH = 4.76 + log(20/2)

And finally calculate the pH:

pH = 5.76

The pH of that acetic acid solution is 5.76.

6 0

3 years ago

Which response includes all the following processes that are accompanied by an increase in entropy? 1) 2SO 2(g) + O 2(g) → SO 3(

Mashutka [201]

Answer: Reaction (1) , (3) and (4) are accompanied by an increase in entropy.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

(1) $2SO_2(g)+O_2(g)\rightarrow SO_3(g)$

3 moles of reactant are changing to 1 mole of product , thus the randomness is increasing. Thus the entropy also increases.

2) $H_2O(l)\rightarrow H_2O(s)$

1 mole of Liquid reactant is changing to 1 mole of solid product , thus the randomness is decreasing. Thus the entropy also decreases.

3) $Br_2(l)\rightarrow Br_2(g)$

1 mole of Liquid reactant is changing to 1 mole of gaseous product , thus the randomness is increasing. Thus the entropy also increases.

4) $H_2O_2(l)\rightarrow H_2O(l)+\frac{1}{2}O_2(g)$

1 mole of Liquid reactant is changing to half mole of gaseous product and 1 mole of liquid product, thus the randomness is increasing. Thus the entropy also increases.

4 0

3 years ago