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Buffer capacity of acidic buffer solutio...

Buffer capacity of acidic buffer solution is maximum when
(1) `P^(H)=P^(k)` (2) [salt ]= [acid ] (3) `p^(K)=7` (4) `[H^(+)]=P^(k)`

A

All are correct

B

(2).(3).(4) are correct

C

(1) and (2) are correct

D

(3) and (2) are correct

Text Solution

Verified by Experts

The correct Answer is:
C
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A certain buffer solution contains equal concentration of X^(-) and HX. K_(b) , for X^(-) is 10^(-10) . Find the pH of buffer.

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_(a) =pK_(a) , or pOH= pk_(b) Which among the following solution will be the most efficient buffer?

Knowledge Check

  • Buffer capacity of acidic buffer solutions is maximum when (a) P^(H) =p^(K) (b) ["Salt"] =["acid "] © p^(K) ,=7 (d) [H^(+)] =p^(K)

    A
    all are correct
    B
    b,c,d, are correct
    C
    a and b are correct
    D
    c and b are correct

  • Higher the amount of acid or base used to produce a definite change of pH in a buffer solutions, 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 +1 In other words any buffer solutions can be used as buffer up 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 , Buffer capacity is maximum when:

    A
    one mole of `NH_4 Cl` is added to one mole of `NH_4 OH`
    B
    one mole of `NH_4 OH ` is added to one mole of HCl
    C
    one mole of `NH_4 Cl ` is added to one mole of NaOH
    D
    one mole of `NaCl ` is added to one mole of NaOH

  • Higher the amount of acid or base used to produce a definite change of pH in a buffer solutions, 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 +1 In other words any buffer solutions can be used as buffer up 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 buffer capacity is equal to :

    A
    `(Delta n)/( Delta pH)`
    B
    ` (pH)/(Delta n) `
    C
    `pK_a =+- 1`
    D
    none of these

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    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_(a) =pK_(a) , or pOH= pk_(b) The bulfer capacity is equal to :

    When a salt reacts with water to form acidic (or ) basic solution the process is called salt hydrolysis. The P^(H) of salt solution can be calculated using the following relation P^(H) =(1)/(2) [P^(k_w) +P^(Ka) + log C ] for salt of weak acid and strong base. P^(H) =(1)/(2) [P^(K_w) -P^(K_a) -log C ] for salt of weak base and strong acid , P^(H) =(1)/(2) [P^(K_w) +P^(K_a) - P^(K_b) ] For a salt of weak acid and weak base where .c. represents the concentration of salt . When a weak acid (or) a weak base is not completly neatralised by strong base (or ) strong acid respectively, then formation of buffer takes places. The P^(H) of buffer solution can be calculated using the following relation P^(H) =P^(Ka) +log "" ((["salt")])/(["Acid "]) , P^(OH) =P^(Kb) +log ""(["salt"])/(["base"]) Answer the following questions using the following data pK_a (CH_3COOH) =4.7447 , pK_b (NH_4OH) =4.7447 , P^(K_W) =14. One mole CH_3 COOH and one mole CH_3 COONa are dissolved in water one litre aqueous solution The P^(H) of the resulting solution will be

    When a salt reacts with water to form acidic (or ) basic solution the process is called salt hydrolysis. The P^(H) of salt solution can be calculated using the following relation P^(H) =(1)/(2) [P^(k_w) +P^(Ka) + log C ] for salt of weak acid and strong base. P^(H) =(1)/(2) [P^(K_w) -P^(K_a) -log C ] for salt of weak base and strong acid , P^(H) =(1)/(2) [P^(K_w) +P^(K_a) - P^(K_b) ] For a salt of weak acid and weak base where .c. represents the concentration of salt . When a weak acid (or) a weak base is not completly neatralised by strong base (or ) strong acid respectively, then formation of buffer takes places. The P^(H) of buffer solution can be calculated using the following relation P^(H) =P^(Ka) +log "" ((["salt")])/(["Acid "]) , P^(OH) =P^(Kb) +log ""(["salt"])/(["base"]) Answer the following questions using the following data pK_a (CH_3COOH) =4.7447 , pK_b (NH_4OH) =4.7447 , P^(K_W) =14. 0.001 M NH_4 Cl aqueous solution has P^(H)

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    The pH of a buffer solution is 4.745 . When 0.044 mole of Ba(OH)_(2) is added to 1 lit. of the buffer , the pH changes to 4.756 . Then the buffer capacity is

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