At 298 K, a certain buffer solution contains equal concentrations of X^- and HX. Kb for X^- is 10-10. What is the pH of…

Solution & Explanation

Related Formula

Kₐ \times Kb = Kw = 10⁻¹⁴

pH = pKₐ + ([X^-])/([HX])

Core Logic

Kₐ = 10⁻¹⁴10⁻¹⁰ = 10⁻⁴ pKₐ = 4

Given $[X^-] = [HX]$ $[X^-] = [HX]$:

pH = 4 + (1) = 4 + 0 = 4

Step 1: Conclusion

pH of the buffer solution is 4.

Pattern Recognition

When conjugate acid and base concentrations are equal, $pH = pKₐ = 14 - pKb$ $pH = pK_a = 14 - pK_b$.

Chapter Mix

Class 11 Chemistry: Ionic Equilibrium

More Ionic Equilibrium Previous-Year Questions

Q48 neet_2024_05_may_morning Solubility Product and pH Calculation

In a qualitative analysis,

Bi³⁺

is detected by appearance of precipitate of

BiO(OH)(s)

. Calculate pH when the following equilibrium exists at 298 K:

BiO(OH)(s) leftharpoons BiO^+(aq) + OH^-(aq), K = 4 \times 10⁻¹⁰

(Given:

2 = 0.3010

)

A. (1) 4.699
B. (2) 8.714
C. (3) 9.301
D. (4) 5.286

Solution

Related Formula

Kₛₚ = s² s = Kₛₚ

[H^+] = (Kw)/([OH^-])

pH = - ₁₀ [H^+]

Core Logic

K = [BiO^+][OH^-] = s² = 4 \times 10⁻¹⁰

s = [OH^-] = 2 \times 10⁻⁵ M

[H^+] = 10⁻¹⁴2 \times 10⁻⁵ = 0.5 \times 10⁻⁹ = 5 \times 10⁻¹⁰ M

pH = - (5 \times 10⁻¹⁰) = 10 - 5 = 9.301

Step 1: Final Calculation

pH = 9 + 2 = 9 + 0.3010 = 9.301

Pattern Recognition

For 1:1 salt equilibrium, $s = \sqrt(K)$ $s = \sqrt{K}$. Then $pOH = - s$ $pOH = -\log s$, $pH = 14 - pOH$ $pH = 14 - pOH$.

Chapter Mix

Class 11 Chemistry: Ionic Equilibrium

Q68 neet_2024_05_may_morning Acid-Base Titration Indicators

Phenolphthalein is used as an indicator for the titration of sodium hydroxide solution against a standard solution of oxalic acid. The colour change that is observed at an alkaline pH close to the equivalence point during this titration is:

A. (1) pinkish red to yellow
B. (2) yellow to pinkish red
C. (3) colourless to pink
D. (4) pink to colourless

Solution

Related Formula

Phenolphthalein: pH < 8.3 (Colourless), pH > 10.0 (Pink)

Core Logic

In acidic/neutral medium before endpoint, phenolphthalein is colourless. Near the equivalence point in alkaline medium, it turns pink.

Step 1: Conclusion

Observed colour change is colourless to pink (Option 3).

Pattern Recognition

Phenolphthalein indicator transition in alkali titrations: Colourless $arrow$ $\rightarrow$ Pink.

Chapter Mix

Class 11 Chemistry: Equilibrium

Q87 neet_2024_05_may_morning Relation between Kp and Kc

Given below are certain reactions. Identify the reaction for which

KP \neq KC

A. (1) $H₂(g) + I₂(g) leftharpoons 2HI(g)$
B. (2) $N₂(g) + O₂(g) leftharpoons 2NO(g)$
C. (3) $N₂(g) + 3H₂(g) leftharpoons 2NH₃(g)$
D. (4) $H₂O(g) + CO(g) leftharpoons H₂(g) + CO₂(g)$

Solution

Related Formula

KP = KC(RT)Δ ng

Core Logic

$KP \neq KC$ $K_P \neq K_C$ whenever $Δ ng \neq 0$ $\Delta n_g \neq 0$. (1) $Δ ng = 2 - 2 = 0 KP = KC$ (2) $Δ ng = 2 - 2 = 0 KP = KC$ (3) $Δ ng = 2 - (1 + 3) = -2 \neq 0 KP \neq KC$ (4) $Δ ng = 2 - 2 = 0 KP = KC$

Step 1: Conclusion

Reaction (3) has $KP \neq KC$ $K_P \neq K_C$.

Pattern Recognition

If sum of gaseous moles on reactant side equals product side, $KP = KC$ $K_P = K_C$.

Chapter Mix

Class 11 Chemistry: Equilibrium

At 298 K, a certain buffer solution contains equal concentrations of $X^-$ $X^-$ and $HX$ $HX$. $Kb$ $K_b$ for $X^-$ $X^-$ is $10⁻¹⁰$ $10^{-10}$. What is the pH of this buffer solution?

Solution & Explanation

Related Formula

Core Logic

Step 1: Conclusion

Pattern Recognition

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At 298 K, a certain buffer solution contains equal concentrations of X^-$X^-$ and HX$HX$. Kb$K_b$ for X^-$X^-$ is 10⁻¹⁰$10^{-10}$. What is the pH of this buffer solution?

Solution & Explanation

Related Formula

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Step 1: Conclusion

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At 298 K, a certain buffer solution contains equal concentrations of $X^-$ $X^-$ and $HX$ $HX$. $Kb$ $K_b$ for $X^-$ $X^-$ is $10⁻¹⁰$ $10^{-10}$. What is the pH of this buffer solution?