Test: PH of Solutions - JEE MCQ

Volume of the original solution = 1 L

pH of the new solution = 2

Volume of the new solution = ?
As per volumetric principle

That means the volume of water added to the original solution of 1 L = 10 −1 = 9 L

so

The beyond the equivalence point can be calculated as follows

before
If units, the before the equivalence point must be At this point if the
reaction is complete, we have only m,mol of HA unreacted,
Hence

K_a of HCN is the order of 10⁻¹⁰ so at this concentration, a of HCN will be negligible.

 which is higher than and
∴ Options , and are wrong.

Let, v of NaOH be added then
No. of mole of salt NaA formed
No. of mole of acid remaining unreacted
for


So

Ammonium benzoate (C₆H₅COONH₄) is the salt weak benzoic acid (C₆H₅COOH,pK_a = 4.25) and weak ammonium hydroxide (NH₄OH, pKb = 4.75).
So, pH of the solution of C₆H₅COONH₄ will be,

As we know, pH = −log[H⁺]
For neutral solution, pH = 7
For acidic solution, pH < 7
For basic solution, pH > 7
Here due to strong base and neutral salt, it's a basic solution. So its pH will be greater than 7.

If escapes, the equilibrium will shift to LHS and concentration will decrease