A weak acid HA has degree of dissociation x. Which option gives the correct expression of pH

Q72 jee_main_2026_24_january_morning Precipitation and pH Dependence

Consider two Group IV metal ions

X²⁺

and

Y²⁺

. A solution containing

0.01 M X²⁺

and

0.01 M Y²⁺

is saturated with

H₂S

. The pH at which the metal sulphide YS will form as a precipitate is ____ (Nearest integer) (Given :

Kₛₚ(XS) = 1 \times 10⁻²²

at

25°C

,

Kₛₚ(YS) = 4 \times 10⁻¹⁶

at

25°C

,

[H₂S] = 0.1 M

in solution,

Kₐ₁ \times Kₐ₂(H₂S) = 1.0 \times 10⁻²¹

,

2 = 0.30

,

3 = 0.48

,

5 = 0.70

)

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

For precipitation: [M²⁺][S²⁻] \geq Kₛₚ

[S²⁻] = Kₐ₁ \cdot Kₐ₂ \cdot [H₂S][H^+]²

Core Logic

For precipitation of $YS(s)$ $YS(s)$: $[Y²⁺][S²⁻] \geq Kₛₚ(YS)$ $[Y^{2+}][S^{2-}] \ge K_{sp}(YS)$ $[S²⁻] \geq 4 \times 10⁻¹⁶0.01 = 4 \times 10⁻¹⁴ M$

Since $H₂S$ $H_2S$ dissociates in water: $H₂S leftharpoons 2H^+ + S²⁻$ $H_2S \rightleftharpoons 2H^+ + S^{2-}$ $[S²⁻][H^+]²[H₂S] = Kₐ₁ \times Kₐ₂ = 1.0 \times 10⁻²¹$

Substitute the $[S²⁻]$ $[S^{2-}]$ threshold: $[S²⁻] = 1.0 \times 10⁻²¹ \times [H₂S][H^+]² \geq 4 \times 10⁻¹⁴$

Step 1: Calculate pH

Using $[H₂S] = 0.1 M$ $[H_2S] = 0.1 \text{ M}$: $10⁻²¹ \times 0.1[H^+]² = 4 \times 10⁻¹⁴$ $10⁻²²[H^+]² = 4 \times 10⁻¹⁴$ $[H^+]² = 10⁻²²4 \times 10⁻¹⁴ = (1)/(4) \times 10⁻⁸ = 0.25 \times 10⁻⁸ = 25 \times 10⁻¹⁰$ $[H^+] = 5 \times 10⁻⁵ M$ $[H^+] = 5 \times 10^{-5} \text{ M}$

$pH = - (5 \times 10⁻⁵) = 5 - 5 = 5 - 0.70 = 4.3$ Rounding to the nearest integer, pH $\approx 4$ $\approx 4$.

Pattern Recognition

Equating the $S²⁻$ $S^{2-}$ concentration required for $Kₛₚ$ $K_{sp}$ with the $S²⁻$ $S^{2-}$ provided by the $H₂S$ $H_2S$ equilibrium directly links solubility product to pH.

Chapter Mix

Class 11 Chemistry: Equilibrium Class 12 Chemistry: Qualitative Analysis

Q55 jee_main_2026_24_january_evening Le Chatelier's Principle and Equilibrium Constant

Consider the following gaseous equilibrium in a closed container of volume "V" at T(K).

P _ 2 (g) + Q _ 2 (g) leftharpoons 2 PQ (g)

2 moles each of

P₂(g)

,

Q₂(g)

and PQ (g) are present at equilibrium. Now one mole each of '

P₂

' and '

Q₂

' are added to the equilibrium keeping the temperature at T(K). The number of moles of

P₂

,

Q₂

and PQ at the new equilibrium, respectively, are -

A. 2.67, 2.67, 2.67
B. 1.21, 2.24, 1.56
C. 1.66, 1.66, 1.66
D. 2.56, 1.62, 2.24

Solution

Core Logic

P₂(g) + Q₂(g) leftharpoons 2PQ(g)

Initially at equilibrium ( $t = teq$ $t = t_{eq}$): 2 mole, 2 mole, 2 mole.

Calculate $Keq$ $K_{eq}$:

Keq = [PQ]²[P₂][Q₂] = 2²2 \cdot 2 = 1

(Since $Δ ng = 0$ $\Delta n_g = 0$, volume $V$ $V$ cancels out in the expression.)

Now 1 mole of each $P₂$ $\mathrm{P}_{2}$ and $Q₂$ $\mathrm{Q}_{2}$ is added. The reaction will move in the forward direction. $t = t'eq$ $t = t'_{eq}$ : Moles of $P₂ = 2 + 1 - x = 3 - x$ $\mathrm{P}_2 = 2 + 1 - x = 3 - x$ Moles of $Q₂ = 2 + 1 - x = 3 - x$ $\mathrm{Q}_2 = 2 + 1 - x = 3 - x$ Moles of $PQ = 2 + 2x$ $\mathrm{PQ} = 2 + 2x$

Step 1: Solve for x

Kc = 1 = (2 + 2x)²(3 - x)(3 - x)

Taking the square root on both sides:

(2 + 2x)/(3 - x) = 1

$2 + 2x = 3 - x$ $2 + 2x = 3 - x$

3x = 1 x = (1)/(3)

Step 2: Final Moles

At the new equilibrium: Moles of $P₂ = 3 - (1)/(3) = (8)/(3) 2.67$ Moles of $Q₂ = 3 - (1)/(3) = (8)/(3) 2.67$ Moles of $PQ = 2 + 2((1)/(3)) = (8)/(3) 2.67$

Pattern Recognition

For reactions where $Δ ng = 0$ $\Delta n_g = 0$, concentration can be directly substituted with moles in the $Kc$ $K_c$ expression because the volume terms cancel entirely.

Chapter Mix

Class 11 Chemistry: Equilibrium

Q56 jee_main_2026_28_january_morning Buffer Solutions

Consider a weak base 'B' of

pKb = 5.699

. 'x' mL of

0.02~M

HCl and 'y' mL of

0.02~M

weak base 'B' are mixed to make

100~mL

of a buffer of pH 9 at

25°C

. The values of 'x' and 'y' respectively are: [Given:

2 = 0.3010

,

3 = 0.4771

,

5 = 0.699

]

A. $x = 11.1, y = 88.9$ $x = 11.1, y = 88.9$
B. $x = 42.7, y = 57.3$ $x = 42.7, y = 57.3$
C. $x = 14.3, y = 85.7$ $x = 14.3, y = 85.7$
D. $x = 85.7, y = 14.3$ $x = 85.7, y = 14.3$

Solution

Related Formula

pOH = pKb + [ SaltBase]

Step 1: Find Active Moles

Reaction: $B + HCl arrow BH^+ + Cl^-$ Moles of HCl = $0.02x$ $0.02x$ mmol Moles of Base B = $0.02y$ $0.02y$ mmol Since a buffer is formed, Base B is in excess. Moles of Salt formed ( $BH^+$ $\mathrm{BH}^+$) = $0.02x$ $0.02x$ Moles of Base remaining = $0.02y - 0.02x$ $0.02y - 0.02x$

Step 2: Apply Henderson-Hasselbalch Equation

Given $pH = 9$ $\mathrm{pH} = 9$, so $pOH = 14 - 9 = 5$ $\mathrm{pOH} = 14 - 9 = 5$.

5 = 5.699 + [(0.02x)/(0.02y - 0.02x)]

-0.699 = ((x)/(y - x))

We know $5 = 0.699$ $\log 5 = 0.699$, therefore $(1/5) = -0.699$ $\log(1/5) = -0.699$.

(x)/(y - x) = (1)/(5)

5x = y - x y = 6x

Step 3: Solve for x and y

Total volume $x + y = 100~mL$ $x + y = 100\mathrm{~mL}$.

x + 6x = 100 7x = 100 x = (100)/(7) \approx 14.3~mL

y = 100 - 14.3 = 85.7~mL

Pattern Recognition

Buffer mixing implies partial neutralization. The ratio of acid volume to base volume defines the logarithmic salt-to-base ratio. Using inverse logs directly maps variables to total volume restrictions.

Chapter Mix

Class 11 Chemistry: Equilibrium

Q75 jee_main_2026_28_january_morning pH of Weak Acids

Consider the dissociation equilibrium of the following weak acid

HA leftharpoons H⁺(aq) + A⁻(aq)

If the

pKₐ

of the acid is 4, then the pH of

10~mM

HA solution is _____. (Nearest integer) [Given : The degree of dissociation can be neglected with respect to unity]

Numerical Answer. Answer: 3 to 3

Solution

Related Formula

pH = (1)/(2) [pKₐ - C]

Step 1: Identify Parameters

$pKₐ = 4$ $\mathrm{pK}_a = 4$

Concentration $C = 10~mM = 10 \times 10⁻³~M = 10⁻²~M$

Step 2: Execute Calculation

pH = (1)/(2) [4 - (10⁻²)]

= (1)/(2) [4 - (-2)]

= (1)/(2) [6] = 3

Pattern Recognition

For weak monoprotic acids where $α ll 1$ $\alpha \ll 1$, the pH is always the arithmetic mean of the $pKₐ$ $\mathrm{pK}_a$ and the negative logarithm of molarity.

Chapter Mix

Class 11 Chemistry: Equilibrium

Q54 jee_main_2026_28_january_evening Le Chatelier Principle And Equilibrium Calculation

Observe the following equilibrium in a 1 L flask.

A(g) leftharpoons B(g)

At T(K), the equilibrium concentrations of A and B are

0.5 M

and

0.375 M

respectively.

0.1 moles

of A is added into the flask and heated to T(K) to establish the equilibrium again. The new equilibrium concentrations (in M) of A and B are respectively.

A. $(1) 0.367, 0.275$ $(1)\text{ 0.367, 0.275}$
B. $(2) 0.53, 0.4$ $(2)\text{ 0.53, 0.4}$
C. $(3) 0.742, 0.557$ $(3)\text{ 0.742, 0.557}$
D. $(4) 0.557, 0.418$ $(4)\text{ 0.557, 0.418}$

Solution

Related Formula

Keq = [B]eq[A]eq

Core Logic

Initial equilibrium: $A leftharpoons B$ $A \rightleftharpoons B$ $[A] = 0.5 M$ $[A] = 0.5\text{ M}$, $[B] = 0.375 M$ $[B] = 0.375\text{ M}$

Keq = (0.375)/(0.5) = 0.75

Now $0.1 mole$ $0.1\text{ mole}$ of $A$ $A$ is added to the 1 L flask. New initial $[A] = 0.5 + 0.1 = 0.6 M$ $[A] = 0.5 + 0.1 = 0.6\text{ M}$. Reaction shifts forward: $A leftharpoons B$ $A \rightleftharpoons B$ Eq: $0.6 - x 0.375 + x$ $0.6 - x \quad 0.375 + x$

Step 1: Solve for x

Keq = 0.75 = (0.375 + x)/(0.6 - x)

0.75(0.6 - x) = 0.375 + x

0.45 - 0.75x = 0.375 + x

$1.75x = 0.075$ $1.75x = 0.075$

x = (0.075)/(1.75) = (3)/(70) \approx 0.043 M

Step 2: Calculate New Concentrations

New concentration of $A = 0.6 - 0.043 = 0.557 M$ $A = 0.6 - 0.043 = 0.557\text{ M}$ New concentration of $B = 0.375 + 0.043 = 0.418 M$ $B = 0.375 + 0.043 = 0.418\text{ M}$

Pattern Recognition

Le Chatelier addition problems require maintaining $Keq$ $K_{eq}$ constant. Set up ICE table after perturbation, equate to initial $Keq$ $K_{eq}$.

Chapter Mix

Class 11 Chemistry: Equilibrium

A weak acid HA$\mathrm{HA}$ has degree of dissociation x$x$. Which option gives the correct expression of pH - pKₐ$\mathrm{pH} - \mathrm{pK}_{\mathrm{a}}$ ?

Solution & Explanation

Related Formula

Step 1: Expressing Concentration

Step 2: Logarithmic Rearrangement

Pattern Recognition

Chapter Mix

Reference Study Guides

More Ionic Equilibrium Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Calculate pH

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Solve for x

Step 2: Final Moles

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Find Active Moles

Step 2: Apply Henderson-Hasselbalch Equation

Step 3: Solve for x and y

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Identify Parameters

Step 2: Execute Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for x

Step 2: Calculate New Concentrations

Pattern Recognition

Chapter Mix

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A weak acid $HA$ $\mathrm{HA}$ has degree of dissociation $x$ $x$. Which option gives the correct expression of $pH - pKₐ$ ?