0

Guest

Courses New

1.Calculate the pH of solution obtained by mixing 10ml of 0.1 M HCl and 40ml of 0.2 M H 2 SO 4. 2.Calculate the pH of solution obtained by mixing 0.1 litre of pH=4 and 0.2 litre of pH=10. 3.Calculate the pH of solution which contains 9.9ml of 1 M HCl and 100ml of 0.1 M NaOH.

1.Calculate the pH of solution obtained by mixing 10ml of 0.1 M HCl and 40ml of 0.2 M H2SO4.
2.Calculate the pH of solution obtained by mixing 0.1 litre of pH=4 and 0.2 litre of pH=10.
3.Calculate the pH of solution which contains 9.9ml of 1 M HCl and 100ml of 0.1 M NaOH.

Grade:10

2 Answers

Arun
25750 Points
6 years ago
1.
H2SO4 dissociates: 
H2SO4 → H+ + HSO4- 
This first dissociation step occurs almost completely which makes H2SO4 a strong acid 
But the second step: 
HSO4- → H+ + SO4 2- occurs to a small extent HSO4 - is a weak acid . Its contrbution to the [H+] will be small - and even further reduced in the presence of the high [H+] from the HCl . 
HCl is a strong acid that dissociuates completely. 

The total [H+] in the final solution is calculated:
Mol H2SO4 in 40mL of 0.2M solution = 40/1000*0.2 = 0.008 mol H2SO4 
This will dissociate to produce 0.008 mol H+ 
Mol HCl in 10mL of 0.1M solution = 10/1000*0.1 = 0.001 mol HCl 
This will dissociate to produce 0.001 mol H+ 

There is therefore a total of 0.009 mol H+ dissolved in 50mL solution = 0.050L 
Molarity of H+ = 0.009/0.050 = 0.18 M 

pH = -log [H+] 
pH = -log 0.18 
pH = 0.74
 
2.
 

ph of strong acid = 4

 

[H+] = 10-4 = molarity of acid

 

in 0.1L , moles of acid = 0.1* 10-4

 

ph of base =10

 

[H+]=10-10

 

[OH-]=10-14/10-10=10-4 = molarity of base

 

in 0.2L , moles of acid = 0.2*10-4

 

now nutralization will occur and acid is completely nutralised by base , at the end base remaining due to higher concentration....

 

now final molariry = remaining moles/total vol = (0.2*10-4-0.1*10-4 ) /0.3

 

                     concentration of base               =10-4/3

 

[H+] = 3*10-14/10-4 =3*10-10

 

ph=-loh[H+] = 10-log3= 9.523

 

3.

NaOH reacts with HCl : 
NaOH + HCl → NaCl + H2O 
1 mol NaOH reacts with 1 mol HCl 

Mol HCl in 9.9mL of 1.0M solution = 9.9/1000*1.0 = 0.0099 mol HCl 
Mol NaOH in 100mL of 0.1M solution = 100/1000*0.1 = 0.01 mol NaOH 
On mixing the 0.0099 mol HCl will be neutralised and 0.01 - 0.0099 = 1*10^-4 mol NaOH dissolved in 109.9mL solution 
Molarity of NaOH solution = (1*10^-4) / 0.1099 = 9.099*10^-4M 
pOH = -log ( 9.099*10^-4) 
pOH = 3.04 
pH = 14.00 - pOH 
pH = 14.00 - 3.04 
pH = 10.96

Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the attached answer to your problem below.

1. H2SO4 dissociates:H2SO4 → H+ + HSO4-
This first dissociation step occurs almost completely which makes H2SO4 a strong acidBut the second step:HSO4- → H+ + SO4 2- occurs to a small extent HSO4 - is a weak acid . Its contribution to the [H+] will be small - and even further reduced in the presence of the high [H+] from the HCl .
HCl is a strong acid that dissociates completely.The total [H+] in the final solution is calculated:
Mol H2SO4 in 40mL of 0.2M solution = 40/1000*0.2
= 0.008 mol H2SO4
This will dissociate to produce 0.008 mol H+Mol HCl in 10mL of 0.1M solution =10/1000*0.1
= 0.001 mol HCl
This will dissociate to produce 0.001 mol H+
There is therefore a total of 0.009 mol H+ dissolved in 50mL solution = 0.050L
Molarity of H+ = 0.009/0.050 = 0.18 M
pH = -log [H+]
pH = -log 0.18pH
= 0.74

2. pH of strong acid = 4
[H+] = molarity of acid = 10-4
In 0.1L, moles of acid = 0.1* 10-4
pH of base =10
[H+] =10-10
OH- = 10-14/10-10 = 10-4 = molarity of base
in 0.2L , moles of acid = 0.2*10-4
now
neutralization will occur and acid is completely neutralised by base , at
the end base remaining due to higher concentration....
now final molariry = remaining moles/total vol = (0.2*10-4 – 0.1*10-4) /0.3 concentration of base =10-4/3
[H+] = 3*10-14/10-4 = 3*10-10
pH = - log[H+] = 10-log3= 9.523
3.
NaOH reacts with HCl :
NaOH + HCl → NaCl + H2O
1 mol NaOH reacts with 1 mol HCl

Mol HCl in 9.9mL of 1.0M solution = 9.9/1000*1.0 = 0.0099 mol HCl
Mol NaOH in 100mL of 0.1M solution = 100/1000*0.1 = 0.01 mol NaOH
On mixing the 0.0099 mol HCl will be neutralised and 0.01 - 0.0099 = 1*10^-4 mol NaOH dissolved in 109.9mL solution
Molarity of NaOH solution = (1*10^-4) / 0.1099 = 9.099*10^-4M
pOH = -log ( 9.099*10^-4)
pOH = 3.04
pH = 14.00 - pOH
pH = 14.00 - 3.04
pH = 10.96

Thanks and regards,
Kushagra

Think You Can Provide A Better Answer ?

Other Related Questions on Physical Chemistry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More