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pOH=7-0.5pKa+0.5 pKb is true for aqueous...

`pOH=7-0.5pK_a+0.5 pK_b` is true for aqueous solution containing which pair of cation and anion :

A

`C_6H_5NH_3^+,CH_3COO^(-)`

B

`NH_4^+,F^(-)`

C

Both (A) and (B)

D

None of these

Text Solution

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The correct Answer is:
To solve the question, we need to analyze the equation given: \[ \text{pOH} = 7 - 0.5 \text{pK}_a + 0.5 \text{pK}_b \] This equation is generally applicable for solutions containing a weak acid and its conjugate base, or a weak base and its conjugate acid. ### Step-by-Step Solution: 1. **Understanding the Components**: - The equation relates pOH to the dissociation constants of a weak acid (pK_a) and a weak base (pK_b). - A weak acid partially dissociates in solution, while a weak base does not fully accept protons. 2. **Identifying the Pairs**: - We need to identify which pairs of cations and anions correspond to weak acids and weak bases. - The options provided are: - A) C₆H₅NH₃⁺ (anilinium ion) and CH₃COO⁻ (acetate ion) - B) NH₄⁺ (ammonium ion) and F⁻ (fluoride ion) 3. **Analyzing Option A**: - **C₆H₅NH₃⁺ (Anilinium ion)**: This is the conjugate acid of aniline (C₆H₅NH₂), which is a weak base. Therefore, C₆H₅NH₃⁺ is a weak acid. - **CH₃COO⁻ (Acetate ion)**: This is the conjugate base of acetic acid (CH₃COOH), which is a weak acid. Therefore, CH₃COO⁻ is a weak base. - Conclusion: This pair consists of a weak acid and a weak base. 4. **Analyzing Option B**: - **NH₄⁺ (Ammonium ion)**: This is the conjugate acid of ammonia (NH₃), which is a weak base. Therefore, NH₄⁺ is a weak acid. - **F⁻ (Fluoride ion)**: This is the conjugate base of hydrofluoric acid (HF), which is also a weak acid. Therefore, F⁻ acts as a weak base. - Conclusion: This pair also consists of a weak acid and a weak base. 5. **Final Conclusion**: - Both pairs (A and B) consist of a weak acid and a weak base, thus satisfying the condition of the equation provided. - Therefore, the correct answer is that both pairs are valid, and the answer is **both A and B are correct**.
To solve the question, we need to analyze the equation given: \[ \text{pOH} = 7 - 0.5 \text{pK}_a + 0.5 \text{pK}_b \] This equation is generally applicable for solutions containing a weak acid and its conjugate base, or a weak base and its conjugate acid. ### Step-by-Step Solution: ...
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Knowledge Check

  • The pK_a of a weak acid (HA) is 4.5 The pOH of an aqueous buffered solution of HA in which 50% of it is ionised is

    A
    `4.5`
    B
    ` 2.5`
    C
    ` 9.5`
    D
    ` 7.0`

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    Pure water is netural in nature [H^(+)]=[OH^(-)]. When this condition is disturbed by changing the concentration of H^(+)or OH^(-), the natural solution changes to acidic {[H^(+)]gt[OH^(-)]} or basic {[H^(+)]lt[OH^(-)]}. This change occurs during salt hydrolysis. pH of salt solution can be calculate using the following relation (i) Salt of weak acid and strong base pH =1/2[pK_(w)+pK_(a)+logC] (ii) Salt of weak base and strong acid pH=1/2[pK_(w)-pK_(b)-logC] (iii) For salt of weak base and strong acid pH=1/2[pK_(w)+pK_(a)-pK_(b)] The pH of buffer can be calculated using t he following formula pH=pK_(a)+log""(["Salt"])/(["Acid"]) pOH=pK_(b)=log""(["Salt"])/(["Base"]) Answer t he following questions when pK_(a)=4.7447 pK_(b)=4.7447 ltb rgt pK_(w)=14 1 mole of CH_(3)COOH is dessolved in water to from 1 litre aqueous solution. The pH of resulting solution will be

    Higher the amount of acid or base used to produce a definite change of pH in a buffer solution,higher will be its buffer capacity. Buffer capacity of solution is maximum under the following conditions: [Salt]= [Acid] (in acid buffer), [Salt] = [Base] (in base buffer) pH of a buffer solution lies in the range given below: pH = pK_(a)pm1 , In other words, any buffer solution can be used as buffer up to two pH units only, depending upon the value of pK_(a) , or pK_(b), . A buffer is said to be efficient when pH =pK_(a) , or pOH= pK_(b) The bulfer capacity is equal to :

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