JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Equilibrium

WhatsApp Email

The pH of a 0.01 M weak acid HX (Kₐ = 4 × 10⁻¹⁰) is found to be 5. Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6. The new concentration of the diluted weak acid is given as x × 10⁻⁴ M. The value of x is ____ (nearest integer)

Answer: 25

Solution

Treating the very weak acid at high dilution as

[H^{+}] = K_{a} C

, the diluted solution at

pH = 6

has

[H^{+}] = 1 0^{- 6} M

. Squaring,

1 0^{- 12} = K_{a} C = (4 \times 1 0^{- 10}) C

, so

C = \frac{1 0 ^{- 12}}{4 \times 1 0 ^{- 10}} = 2.5 \times 1 0^{- 3} M = 25 \times 1 0^{- 4} M

. The value of

x

25

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Integer answer

Practise more: all JEE Main PYQs on Equilibrium →

More JEE Main questions on Equilibrium

The first and second ionization constants of $H_{2} X$ are $2.5 \times 1 0^{- 8}$ and $1.0 \times 1 0^{- 13}$ respectively. The concentration of $X^{2 -}$ in $0.1 M$ $H_{2} X$ solution is $____$ $\times 1 0^{- 15} M$ . (N …
JEE Main 2026View solution →
Which of the following mixture gives a buffer solution with $pH = 9.25$ ?
Given : $pK_{b} (NH_{4} OH) = 4.75$
JEE Main 2026View solution →
Consider the general reaction given below at 400 K
$x A (g) y B (g)$

The values of $K_{p}$ and $K_{c}$ are studied under the same condition of temperature but variation in x and y.
(i) $K_{p} = 85.87$ and $K_{c} = 2.586$ appropriate units
(ii) $K_{p} = 0.862$ and …
JEE Main 2026View solution →
For the following gas phase equilibrium reaction at constant temperature,
$NH X_{3} (g) \frac{1}{2} N X_{2} (g) + \frac{3}{2} H X_{2} (g)$

If the total pressure is $3$ atm and the pressure equilibrium constant ( $K_{p}$ ) is 9 atm, then the degree of dissociation is given as …
JEE Main 2026View solution →
$X_{2} (g) + Y_{2} (g) ⇌ 2 Z (g)$
$X_{2} (g)$ and $Y_{2} (g)$ are added to a 1 L flask and it is found that the system attains the above equilibrium at …
JEE Main 2026View solution →
Consider three metal chlorides $x, y$ and z , where x is water soluble at room temperature, $y$ is sparingly soluble in water at room temperature and z is soluble in hot water. $x, y$ and z are respectively
JEE Main 2026View solution →