Question

1. 100 ml of 0.05 M H2S + 50ml of 0.1 M NaOH

2. 200 ml of 0.1 M Na2CO3 + 100 ml of 0,365% HCl

3. 50ml of 0.1M Na3PO4 + 100ml of 0.05 M H2SO4

4. 80 ml of 0.01M HNO2 + 20ml of 0.05 M HNO3

5. 20 ml of 0.05 M H2C2O4 + 200 ml of 0.01 M KOH

6. 100 ml of 0.05 M H2S + 100 ml of 0.1 M NaOH

7. 50 ml of 0.01 M NaOH + 50 ml of 0.02 M NH3

8.100 ml of 0.01 M NH3 + 10 ml of 0.365 % HCl

9. 100 ml of 0.2 M Na2CO3 + 100 ml of 0.5 % HCl

10. 100 ml of 0.05 M H2CO3 + 50 ml of 0.1 M NaOH

11. 50 ml of 0.1M Na3PO4 + 100 ml of 0.05 M HNO3

12. 20 ml of 0.025 M H2C2O4 + 100 ml of 0.01 M KOH

13. 100 ml of 0.01 M HCl + 10 ml of 0.5 % NH3

How to find pH of solutions

Answer 1

1.Ans :- 100 ml of 0.05 M H₂S + 50 ml of 0.1 M NaOH :-

No. of moles of H₂S = Molarity x Volume in L

= 0.05 M x 0.100 L

= 0.005 mol

Also,

No. of moles of NaOH = Molarity x Volume in L

= 0.1 M x 0.050 L

= 0.005 mol

ICE table is :

.............................H₂S..................+....................NaOH ----------------------> NaHS .................+....................H₂O

Initial....................0.005 mol................................0.005 mol.........................0.0 mol.......................................

Change...............-0.005 mol...............................-0.005 mol.........................+0.005 mol...................................

Final.....................0.0 mol.....................................0.0 mol...............................0.005 mol.........................................

Total volume = 100 mL + 50 mL = 150 mL = 0.150 L

Molar concentration of NaHS = No. of moles / Volume in L

= 0.005 mol / 0.150 L

= 0.0333 mol/L

Since, NaHS is a salt of strong base NaOH and weak acid H₂S and its pH can be calculated by using the formula :

pH = 1/2[14 + pK_a + log C]

pH = 1/2 [14 + 6.99 + log 0.033 ]

pH = 1/2 [14 + 6.99 - 1.48 ]

pH = 1/2 x 19.51

pH = 9.75

Therefore, pH = 9.75