N Awasthi Physical Chemistry
10^−5 M HCI solution at 25^∘C is dilluted 1000 times. The pH of the resultant solution will.
12
Nov
10^−5 M HCI solution at 25^∘C is dilluted 1000 times. The pH of the resultant solution will. 10^−5 M HCI solution at 25^∘C is dilluted 1000 times. The pH of the resultant solution will. November 12, 2020 Category: Chapter 6 - Ionic Equilibrium (Level 1 and Level 2) , N Awasthi Physical Chemistry ,
pH of 10^-6 M HCL(aq) is :
12
Nov
pH of 10^-6 M HCL(aq) is : pH of 10^-6 M HCL(aq) is : November 12, 2020 Category: Chapter 6 - Ionic Equilibrium (Level 1 and Level 2) , N Awasthi Physical Chemistry ,
Calcium hydroxide is a strong base. Compute [Ca^2+] and [OH−] “for” a solution that is prepared by dissolving 0.60g of Ca(OH)2 in enough water to make a 1500 mL of solution. [Atomic mass : Ca=40,O=16,H=1]
12
Nov
Calcium hydroxide is a strong base. Compute [Ca^2+] and [OH−] “for” a solution that is prepared by dissolving 0.60g of Ca(OH)2 in enough water to make a 1500 mL of solution. [Atomic mass : Ca=40,O=16,H=1] Calcium hydroxide is a strong base. Compute [Ca^2+] and [OH−] "for" a solution that is prepared by dissolving 0.60g of [...]
pH of a strong diprotic acid (H2A) at concentrations: (i) 10^−4 M, (ii) 10^−4 N are respectively:
12
Nov
pH of a strong diprotic acid (H2A) at concentrations: (i) 10^−4 M, (ii) 10^−4 N are respectively: (ii) 10^−4 N are respectively: pH of a strong diprotic acid (H2A) at concentrations: (i) 10^−4 M November 12, 2020 Category: Chapter 6 - Ionic Equilibrium (Level 1 and Level 2) , N Awasthi Physical Chemistry ,
The pH of a solution of H2SO4 is 1. Assuming complete ionisation, find the molarity of H2SO4 solution :
12
Nov
The pH of a solution of H2SO4 is 1. Assuming complete ionisation, find the molarity of H2SO4 solution : find the molarity of H2SO4 solution : The pH of a solution of H2SO4 is 1. Assuming complete ionisation November 12, 2020 Category: Chapter 6 - Ionic Equilibrium (Level 1 and Level 2) , N Awasthi [...]
To a 10 ml of 10^-3 nh2so4 solution water has been added to make the total volume of one litre. Its pOH would be :
12
Nov
To a 10 ml of 10^-3 nh2so4 solution water has been added to make the total volume of one litre. Its pOH would be : To a 10 ml of 10^-3 nh2so4 solution water has been added to make the total volume of one litre. Its pOH would be : November 12, 2020 Category: Chapter [...]
Number of equivalents of HCl present in 100 mL of its solution whose pH is 4:
12
Nov
Number of equivalents of HCl present in 100 mL of its solution whose pH is 4: Number of equivalents of HCl present in 100 mL of its solution whose pH is 4: November 12, 2020 Category: Chapter 6 - Ionic Equilibrium (Level 1 and Level 2) , N Awasthi Physical Chemistry ,
Equal volumes of two HCl solutions of pH=3 and pH=5 were mixed. What is the PH of the resulting solution ?
12
Nov
Equal volumes of two HCl solutions of pH=3 and pH=5 were mixed. What is the PH of the resulting solution ? Equal volumes of two HCl solutions of pH=3 and pH=5 were mixed. What is the PH of the resulting solution ? November 12, 2020 Category: Chapter 6 - Ionic Equilibrium (Level 1 and Level [...]
pOH of 0.002 M HNO3 is : (a) 11 + log2
12
Nov
pOH of 0.002 M HNO3 is : (a) 11 + log2 pOH of 0.002 M HNO3 is : (a) 11 + log2 November 12, 2020 Category: Chapter 6 - Ionic Equilibrium (Level 1 and Level 2) , N Awasthi Physical Chemistry ,
A solution has a pH=9. It is 1000 times more basic than the original solution. What was the pH of the original solution?
12
Nov
A solution has a pH=9. It is 1000 times more basic than the original solution. What was the pH of the original solution? A solution has a pH=9. It is 1000 times more basic than the original solution. What was the pH of the original solution? November 12, 2020 Category: Chapter 6 - Ionic Equilibrium [...]