JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Equilibrium

Only litre buffer solution was prepared by adding 0.10 mol each of

NH_{3}

and

NH_{4} Cl

in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is

____

\times 1 0^{- 2}

,(Nearest integer)
(Given :

pK_{b}

NH_{3} = 4.745

and

lo g_{10} 3 = 0.477

)

Answer: 48

Solution

This is a base buffer, for which

pOH = p K_{b} + lo g \frac{[ N H _{4}^{+} ]}{[ N H _{3} ]}

. At the start the two components are equal, so the log term vanishes,

pOH = 4.745

and

pH = 9.255

. Adding

0.05

mol of HCl turns an equal amount of ammonia into ammonium, leaving

[N H_{3}] = 0.05

mol and

[N H_{4}^{+}] = 0.15

mol. Now

pOH = 4.745 + lo g 3 = 4.745 + 0.477 = 5.222

, so

pH = 8.778

. The pH falls by

9.255 - 8.778 = 0.477

, which is

47.7 \times 1 0^{- 2}

, and to the nearest integer the change in pH is

48 \times 1 0^{- 2}

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Integer answer