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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.

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3 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
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3 years ago 610 Points
```							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 H2SO4This will dissociate to produce 0.008 mol H+Mol HCl in 10mL of 0.1M solution =10/1000*0.1= 0.001 mol HClThis will dissociate to produce 0.001 mol H+There is therefore a total of 0.009 mol H+ dissolved in 50mL solution = 0.050LMolarity of H+ = 0.009/0.050 = 0.18 MpH = -log [H+]pH = -log 0.18pH= 0.742. pH of strong acid = 4[H+] = molarity of acid = 10-4In 0.1L, moles of acid = 0.1* 10-4pH of base =10 [H+] =10-10OH- = 10-14/10-10 = 10-4 = molarity of basein 0.2L , moles of acid = 0.2*10-4nowneutralization will occur and acid is completely neutralised by base , atthe 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*10-10pH = - log[H+] = 10-log3= 9.5233.NaOH reacts with HCl :NaOH + HCl → NaCl + H2O1 mol NaOH reacts with 1 mol HClMol HCl in 9.9mL of 1.0M solution = 9.9/1000*1.0 = 0.0099 mol HClMol NaOH in 100mL of 0.1M solution = 100/1000*0.1 = 0.01 mol NaOHOn mixing the 0.0099 mol HCl will be neutralised and 0.01 - 0.0099 = 1*10^-4 mol NaOH dissolved in 109.9mL solutionMolarity of NaOH solution = (1*10^-4) / 0.1099 = 9.099*10^-4MpOH = -log ( 9.099*10^-4)pOH = 3.04pH = 14.00 - pOHpH = 14.00 - 3.04pH = 10.96Thanks and regards,Kushagra
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6 months ago
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