Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

Get extra R 520 off
USE CODE: MOB20

```				   how find ph values
```

6 years ago

Share

```										Hi

pH

pOH

pH is a measure of the hydrogen ion concentration, [H+]

pOH is a measure of the hydroxide ion concentration, [OH-]

pH is calculated using the following formula: pH = -log10[H+]

pOH is calculated using the following formula: pOH = -log10[OH-]

Example 1:
Find the pH of a 0.2mol L-1 (0.2M) solution of HCl

Write the balanced equation for the dissociation of the acid HCl -----> H+(aq) + Cl-(aq)
Use the equation to find the [H+]: 0.2 mol L- HCl produces 0.2 mol L-1 H+ since HCl is a strong acid that fully dissociates
Calculate pH: pH = -log10[H+] pH = -log10[0.2] = 0.7

Example 1:
Find the pOH of a 0.1mol L- (0.1M) solution of NaOH

Write the balanced equation for the dissociation of the alkaliNaOH -----> OH-(aq) + Na+(aq)
Use the equation to find the [OH-]:0.1 mol L-1 NaOH produces 0.1 mol L-1 OH- since NaOH is a strong alkali that fully dissociates
Calculate pOH: pOH = -log10[OH-]pOH = -log10[0.1] = 1

Example 2:
Find the pH of a 0.2 mol L-1 (0.2M) solution of H2SO4

Write the balanced equation for the dissociation of the acid H2SO4 -----> 2H+(aq) + SO42-(aq)
Use the equation to find the [H+]: 0.2 mol L-1 H2SO4 produces 2 x 0.2 = 0.4 mol L-1 H+ since H2SO4 is a strong acid that fully dissociates
Calculate pH: pH = -log10[H+]pH = -log10[0.4] = 0.4

Example 2:
Find the pOH of a 0.1mol L-1 (0.1M) solution of Ba(OH)2

Write the balanced equation for the dissociation of the alkali:Ba(OH)2 -----> 2OH-(aq) + Ba2+(aq)
Use the equation to find the [OH-]: 0.1mol L-1 Ba(OH)2 produces 2 x 0.1 = 0.2 mol L-1 OH- since Ba(OH)2 is a strong alkali that fully dissociates
Calculate pOH: pOH = -log10[OH-] pOH = -log10[0.2] = 0.7

Hydrogen ion concentration, [H+], can be calculated using the following formula: [H+] = 10-pH

Hydroxide ion concentration, [OH-], can be calculated using the following formula: [OH-] = 10-pOH

Example:
Find the [H+] of a nitric acid solution with a pH of 3.0pH= 3.0 [H+] = 10-pH [H+] = 10-3.0 = 0.001mol L-1
You can check this answer by using the calculated value [H+] in the equation for pH to make sure you arrive at the original pHpH = -log10[H+] pH = -log10[0.001] = 3We get the same value for pH using the calculated value for [H+], so the calculated value for [H+] is correct.

Example:
Find the [OH-] of a sodium hydroxide solution with a pOH of 1pOH = 1 [OH-] = 10-pOH [OH-] = 10-1 = 0.1 mol L-1
You can check this answer by using the calculated value [OH-] in the equation for pOH to make sure you arrive at the original pOHpOH = -log10[OH-]pOH = -log10[0.1] = 1We get the same value for pOH using the calculated value for [OH-], so the calculated value for [OH-] is correct.

pH + pOH = 14

Example A(1):
Find the pH of a solution of sodium hydroxide that has a pOH of 2
pH = 14 - pOHpH = 14 - 2 = 12

Example B(1):
Find the pOH of a solution of hydrochloric acid that has a pH of 3.4
pOH = 14 - pHpOH = 14 - 3.4 = 10.6

Example A(2):
Find the [H+] in a solution of sodium hydroxide that has a pOH of 1

Calculate the pHpH = 14 - pOHpH = 14 - 1 = 13
Calculate [H+][H+] = 10-pH[H+] = 10-13 = 10-13mol L-1

Example B(2):
Find the [OH-] of a sulfuric acid solution with a pH of 3

Calculate the pOHpOH = 14 - pHpOH = 14 - 3 = 11
Calculate [OH-][OH-] = 10-pOH[OH-] = 10-11 = 10-11 mol L-1

Example A(3):
Find the pH of 0.2mol L-1 sodium hydroxide

Write the equation for the dissociation of NaOH:NaOH -----> Na+(aq) + OH-(aq)
Use the equation to find [OH-]:0.2mol L-1 NaOH produces 0.2mol L-1 OH- since NaOH is a strong base that fully dissociates
Calculate the pOH:pOH = -log10[OH-]pOH = -log10[0.2] = 0.7
Calculate pH:pH = 14 - pOHpH = 14 - 0.7 = 13.3

Example B(3):
Find the pOH of 0.2mol L-1 sulfuric acid

Write the equation for the dissociation of H2SO4:H2SO4 -----> 2H+(aq) + SO42-(aq)
Use the equation to find [H+]:0.2 mol L-1 H2SO4 produces 2 x 0.2 = 0.4 mol L-1 H+ since H2SO4 is a strong acid that fully dissociates
Calculate the pH:pH = -log10[H+]:pH = -log10[0.4] = 0.4
Calculate pOH:pOH = 14 - pHpOH = 14 - 0.4 = 13.6

Thanks
Anurag Kishore
```
6 years ago

# Other Related Questions on Physical Chemistry

what is meant by entropy..........................................................?

@ sharda entropy of a system is actually the measurement of degree of randomess ordisorderliness of a system . it is a thermodynamic property and represented by a letter S . the entropyof...

 Umakant biswal 4 months ago

entropy is the sudden change in randomness of the system due to may be from exapnding, contracting, collision etc. small example → take a glass of water and Eno. eno prsesnt in packed...

 cris gayle 4 months ago

entropy is the sudden change in randomness of the system due to may be from exapnding, contracting, collision etc. small example → take a glass of water and Eno. eno prsesnt in packed paper...

 Vikas TU 4 months ago
In the chapter Solid State Class 12, how to classify any solid or elements on the basis of their bonds i.e. covalent bond, molecular bond etc ???

Depending upon the type of bonding present , Solids can be divided as : 1) Covalent Solids 2) Ionic solids 3) Molecular solids 4) Metallic solids

Ok , thats the same thing i told , u just need to see the configuration of examples . i am giving that below potassium sulphate – ionic solid tin – mettalic solid benzene – molecular (non...

 Umakant biswal 6 months ago

@ shreyas On the besic of bond present there are no such classification as tetraphosphorous decosides , and aluminium phosphates . Besic of the bond there are 4 types of classification 1-...

 Umakant biswal 6 months ago
what is the oxidation state of sulphur in H 2 S 2 O 8, Na 2 S 4 O 8 and Na 2 S 2 O 3.

@ bhavika in H2S2O8, since 2 oxygen is forming the peroxide linkage here , so, the o.n of sulpher will be +6 . and in the 2 nd case NA2S4O8 , in thaat case take sulpher o.n to be x after...

 Umakant biswal 6 months ago

You no longer have to wait desperately for someone to help resolve your doubt. You can chat with IITians live, 24/7 (even at 3AM!) and get your doubt resolved instantly. Try the HashLearn...

 Ankit 6 months ago

u need to remember that which structure have peroxide linkage and which structure have not , there are only some exception with peroxide linkage , go through the ncert redox reaction chapter...

 Umakant biswal 6 months ago
the hybridization of phosphorus in PCl 5 (in solid state ) is 1)sp 3 2)sp 3 d 2 3)sp 3 d 4)both 1 and 2

sp3d is the right answer. In excited state, intermixing of a 3s, three 3p and one 3d orbitals gives five half filled sp 3 d hybrid orbitals, which are arranged in trigonal bipyramidal...

 dolly bhatia one month ago

@ krishna option no 3 turns out to be the correct option , here , the orbitals are filled by electrons of five cl atom . there are 5 p-cl sigma bond , three in one plane and make an angle of...

 Umakant biswal 3 months ago

i hope to no.3 because phosphours atom contain in last subshell is 5 electron and not contain a lone pair and here is 5 chlorine atom .then we add up eletron and we get a 10 and we devided...

 sahabudin ali 2 months ago

It should be option 4 as PCl5 exists in ［PCl4］+ and ［PCl6］- in solid state....as in first case it is Sp3 and that in second it is Sp3d2

 6 days ago
sir iam a iiit students in RAGIV GANDHI UNIVERSITY OF TECHNOLOGIES, there is a semister it infront of me in 10 days.i don’t know any thing about chemistry.it is a ncert syllubus.please send...

@ badugu they are ncert books topic , i am mentioning some imp chapter name below 1- COORDINATION COMPOUND ( PART 2 CHEMI ) 2- GENERAL PRINCIPLES OF ORGANIC CHEMISTRY ( 11TH 2ND PART ) 3-...

 Umakant biswal one month ago

Dear Vijay, Go through these topics first as atmost importat ones: Aldehyde and Ketones Biomolecules Alcohols and ethers Coordination Compounds Solid states Electrrochemistry Solutions D and...

 Vikas TU one month ago

@ badugu , kindly be specific in asking the ques , earlier u told that u have ncert syllabus , and i have given the chapters from ncert textbooks only . kindly conform which publication u...

 Umakant biswal one month ago
How many unpaired electrons are present in superoxide ion?

But the answer is 3.. As in molecular orbital theory the unpaired electrons in oxygen molecule is 2.. It accepts 1 electron to form superoxide ion.. That electron instead of pairing with 2...

 Suraj Singh 17 days ago

And now I am agree with my solution.. You should check it once.... .. And thanq for the concern... .....

 Suraj Singh 17 days ago

It`s unfortunately wrong..................................... ................. NOť less than 100 char

 Suraj Singh 18 days ago
View all Questions »

• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: R 15,000
• View Details
Get extra R 3,000 off
USE CODE: MOB20

Get extra R 520 off
USE CODE: MOB20

More Questions On Physical Chemistry