All categories

3 years ago

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).

2 answers:

8 0

a) before addition of any KOH :

when we use the Ka equation & Ka = 4 x 10^-8 :

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

= 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

= 9.2 x 10^-5 M

when PH = -㏒[H+]

PH = -㏒(9.2 x 10^-5)

= 4

b)After addition of 25 mL of KOH: this produces a buffer solution

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]

first, we have to get moles of HCIO= molarity * volume

=0.21M * 0.05L

= 0.0105 moles

then, moles of KOH = molarity * volume

= 0.21 * 0.025

=0.00525 moles

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L = 0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

= 0.00525 / 0.075

=0.07 M

and molarity of KCIO = moles KCIO / total volume

= 0.00525 / 0.075

= 0.07 M

and when Ka = 4 x 10^-8

∴Pka =-㏒Ka

= -㏒(4 x 10^-8)

= 7.4

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume

= 0.21 M * 0.05L

= 0.0105 moles

then moles KOH = molarity * volume
= 0.22 M* 0.035 L

=0.0077 moles

∴ moles of HCIO remaining = 0.0105 - 0.0077= 8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume

= 8 x 10^-5 / 0.085

= 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

= 0.0077M / 0.085L

= 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

= 0.0105mol / 0.1 L

= 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

= 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO]

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

= 3.8
∴PH = 14- POH

=14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M

V2 is the excess volume added of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

= -㏒0.02

= 1.7

∴PH = 14- POH

= 14- 1.7

= 12.3

natka813 [3]3 years ago

4 0

PH = 7.58

<h2>Further Explanation </h2>

KOH will interact with HClO to provide KClO. you may have an answer that contains unreacted HClO and KOH. this is often a solution.

<h3>Equation: </h3>

HClO + KOH → KClO + H2O

HClO reacts with KOH in a very 1: 1 molar ratio

Mole of acid in 50 ml of an answer of 0.150 m = 50/1000 * 0,150 = 0.0075 mole of acid

Mole of KOH in 30 ml of 150 M solution = 30/1000 * 0,150 = 0.0045 mole of KOH

This reacts to provide 0.0045 moles of KClO and there are 0.0030 moles of HClO unreacted

The volume of the answer = 50 ml + 30 ml = 80 ml = 0.080 l

Deposition of acid in solution = 0.0030 / 0.080 = 0.0375M

KClO deposit in solution = 0.0045 / 0.080 = 0.0562 m

<h3>Using the Henderson-Hasselbalch equation, we will calculate pH; </h3>

pKa HClO = -log (4.0 * 10 ^ -8) = 7.40

PH = pKa + log ([KClO] / [HClO])

PH = 7.40 + notes (0.0562 / 0.0375)

PH = 7.40 + 1.50 notes

PH = 7.40+ 0.18

PH = 7.58

Learn More

Mole of Acid brainly.com/question/9465562

HClO brainly.com/question/12355703

Details

Grade: College

Subject: Chemistry

Keyword: mole, acid, HClO

You might be interested in

A sample of gasoline has a mass of 595.0 lb and a volume of 14.0 ft^3. What is its specific gravity?

Sedbober [7]

Explanation:

The given data is as follows.

mass of gasoline = 595.0 lb

volume of gasoline = $14.0 ft^{3}$

As it is known that density is the amount of mass of a substance divided by its volume.

So, density of gasoline will be as follows.

Density = $\frac{mass}{volume}$

= $\frac{595.0 lb}{14.0 ft^{3}}$

= 42.5 $lb/ft^{3}$

As, 1 lb = 0.4536 kg and 1 ft = 0.3048 m. Now, putting these values into the above equation (1) as follows.

Density = 42.5 $lb/ft^{3}$

= $42.5 \times \frac{0.4536 kg}{(0.3048 m)^{3}}$

= 680.792 $kg/m^{3}$

= 680.8 $kg/m^{3}$ (approx)

Density of water is 1000 $kg/m^{3}$ . To measure specific gravity of gasoline, the formula will be as follows.

specific gravity of gasoline = $\frac{\text{density of gasoline}}{\text{density of water}}$

= $\frac{680.8 kg/m^{3}}{1000 kg/m^{3}}$

= 0.681

Thus, we can conclude that the specific gravity of gasoline is 0.681.

4 0

4 years ago

How many atoms are present in 0.0845 moles of tungsten?

klasskru [66]

1 mole tungsten ------------- 6.02 x 10²³ atoms
0.0845 moles ---------------- ( atoms tungsten )

atoms tungsten = 0.0845 x ( 6.02 x 10²³) / 1

atoms tungsten = 5.08 x 10²² / 1

= 5.08 x 10²² atoms of tungsten

hope this helps!

6 0

4 years ago

Olivia bought new gym shoes to play volleyball because she kept slipping when she ran in her old shoes. How will the new soles h

HACTEHA [7]

Maybe her old shoes had soft worn out bottoms and she slips in them. So her new shoes had more grip than her old ones so they kept her from falling.

6 0

3 years ago

Read 2 more answers

Jane is sliding down a slide. What kind of motion is she demonstrating?

Alex Ar [27]

Im pretty sure it would be kinetic <span />

8 0

4 years ago

Clear glass, water, and air are examples of what kind of material?

irina1246 [14]

Clear glass, water, and air are examples of transparent material. They can be seen through with no extra factors or complications. Therefore, your answer would be D. transparent.

6 0

4 years ago

Read 2 more answers