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Question
Calculate the pH of solution obtained by mixing 10ml of 0.2M HCL and 40ml of 0.1M H2SO4.
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Solution
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
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
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