# Please solve: 64. Calculate the pH of a solution which contains 9.9 ml of 1 M HCI and 100 ml of 0.1 M NaOH.  65. Calculate the pH of a solution by mixing 0.1 litre of pH - 4 and 0.2 litre of pH = 10. 65. Calculate the pH of a solution obtained by mixing 10 ml of 0.1 M HCI and 40 ml of 0.2 M H2SO4-

Dear Student,

64.
NaOH + HCl → NaCl + H2
1 mol NaOH reacts with 1 mol HCl
Mole of  HCl in 9.9mL of 1.0M solution = 9.9/1000  x 1.0 = 0.0099 mol HCl
Mole of  NaOH in 100mL of 0.1M solution = 100/1000 x 0.1 = 0.01 mol NaOH
On mixing the 0.0099 mol HCl will be neutralised and 0.01 - 0.0099 = 1 x 10-4 mol
1 x 10-4 mol of NaOH dissolved in 109.9mL solution.
Molarity of NaOH solution = (1 x 10-4) / 0.1099 = 9.099 x 10-4
pOH = -log ( 9.099 x 10-4
pOH = 3.04
pH = 14.00 - pOH
pH = 14.00 - 3.04
pH = 10.96

65.
Concentration in pH 4 solution = 0.1 x 10-4
Concentration in pH 10 solution = 0.2 x 10-10
[OH-]=10-14/10-10=10-4 = molarity of base in 0.2L , moles of acid = 0.2 x 10-4
Now neutralization will happen and acid is completely neutralised by base.
Final molarity = remaining moles/total vol = (0.2 x 10-4  -  0.1x 10-4) /0.3
Concentration of base =10-4/3[H+] = 3 x 10-14/10-4 =3 x 10-10
pH=-log[H+] = 10 - log3= 9.523
Concentration of base =10-4/3[H+] = 3 x 10-14/10-4 =3 x 10-10
pH = -log[H+] = 10 - log3 = 9.523

66.
H2SO4 → H+ + HSO4-
HSO4- → H+ + SO42-
Mole of  H2SO4 in 40mL of 0.2M solution = 40/1000 x 0.2 = 0.008 mol H2SO
Mole of HCl in 10mL of 0.1M solution = 10/1000 x 0.1 = 0.001 mol HCl
Therefore total of 0.009 mol H+ is dissolved in 50mL solution = 0.050L
Molarity of H+ = 0.009/0.050 = 0.18 M
pH = -log [H+] = -log 0.18 = 0.74.

Regards,

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