1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ioda
3 years ago
11

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).

Chemistry
2 answers:
Degger [83]3 years ago
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
Which of the following is a heterogeneous mixture
jeka57 [31]
Soil. 
Heterogeneous means is can be separated and soil can be taken apart and separated according to its contents. (Rocks, dirt, grass). <span />
7 0
3 years ago
Determine the mass of water in kg, if 4300 cal of energy is placed in water, resulting in a temperature change to 101.0 oC from
Airida [17]

Answer:

0.5059kg

Explanation:

The heat absorbed for the water is determined using the equation:7

Q = C×m×ΔT

<em>Where Q is heat absorbed (4300cal)</em>

<em>C is specific heat (1cal/g°C)</em>

<em>m is the mass in grams</em>

<em>ΔT is change in °C (101.0°C - 92.5°C = 8.5°C)</em>

<em />

Replacing:

4300cal = 1cal/g°C×m×8.5°C

505.9g = m

In kg, the mass of water is:

<h3>0.5059kg</h3>

<em />

6 0
3 years ago
A chemical bond results from the mutual attrac- tion of the nuclei for
aleksandrvk [35]

for what????? complete the question

8 0
2 years ago
Read 2 more answers
At 25 °C, what is the hydroxide ion concentration, [OH−] , in an aqueous solution with a hydrogen ion concentration of [H+]=4.8×
algol [13]

[OH⁻]= 1 x 10⁻¹⁴ : 4.8 x 10⁻⁴ = 2.083 x 10⁻¹¹

3 0
1 year ago
Write a balanced equation formed when the following elements react with oxygen:a)Zinc
Ainat [17]

Explanation:

a) when zinc burnt in oxygen.

2Zn + O2 -----∆-----> 2ZnO(black residue)

b) when carbon burnt in oxygen.

C+O2----∆---> CO2.

c) when sulphur burnt in oxygen.

S+O2-----∆-----> SO2.

d) when Calcium burnt in oxygen.

2Ca+O2-----∆-----> 2CaO(black residue)

e) when Magnesium burnt in oxygen.

2Mg+O2-----∆----> 2MgO.

f) when sodium burnt in oxygen.

4Na+O2----∆-----> 2Na2O.

hope all these reactions help you.

4 0
2 years ago
Other questions:
  • Which of the following statements is NOT true?
    8·1 answer
  • When considering different hypotheses, usually the ____ one which can account for the ____ is the correct one. |
    11·2 answers
  • An online wave simulator created these four waves. Which wave has the lowest frequency?
    10·2 answers
  • What mass of sulfur would have precisely 4.7x10^22 atoms of sulfur?
    9·2 answers
  • From his experiments, J. J. Thomson concluded that
    7·2 answers
  • (a) 4.64 x 1024 atoms Bi into grams
    8·1 answer
  • 17 POINTS (SERIOUS ANSWER ASAPP)
    9·1 answer
  • Consider the combustion of methane, CH4 :
    5·1 answer
  • Where can you change the atom/molecules used in the simulation? Which atoms/molecules are
    7·1 answer
  • 1. When conducting this experiment, some procedures call for heating the substance several times and recording the mass after ea
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!