Kushagra Madhukar
Last Activity: 4 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
