Question

How to find pH if solutions:
50 ml of 0.1M Na3PO4     + 100 ml of 0.05 M HNO3

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

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

Answer 1

1.Sol :-

(a). Number of moles of Na3PO4 or PO43- = Molarity x Volume in L

= 0.1 M x 0.050 L

= 0.005 mol

Similarly,

Number of moles of HNO3 = Molarity x Volume in L

= 0.05 M x 0.100 L

= 0.005 mol

ICF table is :

....................................PO43- ......................+....................H+ --------------------------> HPO42-

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

Molar concentration of HPO42- = Number of moles / Volume in L

= 0.005 mol / 0.150 L

= 0.033 mol/L

Now,

ICE table of HPO42- is :

...................................HPO42- <-----------------------------------> PO43- .........................+.............................H+

Initial (I)...........................0.033 M.................................................0.0 M..........................................................0.0 M

Change (C)..........................-α.......................................................+α..................................................................+α

Equilibrium (E)................(0.033-α) M.............................................α M..............................................................α M

Here, α = Degree of dissocitaion

Expression of acid dissociation constant i.e. Ka (is equal to the ratio of molar concentration of products to the molar concentration of reactants raise to power their stoichiometric coefficient at equilibrium stage).

Ka value of HPO42- = 4.8 × 10−13

Ka = [PO43-].[H+]/[HPO42]

4.8 × 10−13 = α2/(0.033-α)

If α <<<0.033, then neglect α

α2 = 0.033 x 4.8 x 10-13

α = (1.58 x 10-14)1/2

α = 1.26 x 10-7

So, [H+] = α = 1.26 x 10-7 M

As. pH = - log [H+]

So,

pH = - log 1.26 x 10-7 M

= 6.90

Hence, pH of the resulting solution = 6.90