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Equilibrium Test 6

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Equilibrium Test 6
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  • Question 1
    1 / -0

    The conjugate acid of NH2- is

    Solution

    Conjugate acid of a base is formed by adding H+ to it
    NH2- + H+ → NH3
     

  • Question 2
    1 / -0

    Out of the following, amphiprotic species are

    I : HPO32-

    II OH-

    III H2PO4-

    IV HCO3-

    Solution

    Amphiprotic : can accept and Release H+
    Only H2PO–4 & HCO–3

  • Question 3
    1 / -0

    pH of an aqueous solution of NaCl at 85°C should be

    Solution

    NaCl Solution : pH is the, pH   of water.

    pH at 25°C <7

  • Question 4
    1 / -0

    1 c.c. of 0.1N HCl is added to 99 CC solution of NaCl. The pH of the resulting solution will be

    Solution

    Volume of resulting solution = 100 ml

  • Question 5
    1 / -0

    10 ml of  is mixed with 40 ml of . The pH of the resulting solution is

    Solution


    [H+] = 10–2
    pH = 2.

  • Question 6
    1 / -0

    If pKb for fluoride ion at 25°C is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is :

    Solution

    H+  H+ + F–
    pKw = pKa + pKb
    [For conjugate Acid-Base]
    ⇒ pKa = 14 – 10.87 = 3.17
    Ka = 6.76 × 10–4

  • Question 7
    1 / -0

    The pH of an aqueous solution of 1.0 M solution of a weak monoprotic acid which is 1% ionised is:

    Solution

    Concentration of solution =C=1M
    Dissociation is α=0.01
    After dissociation, the concentration of H+ ions will be Cα=1×0.01=10−2
    pH=−log[H+]=−log(10−2) = +2

  • Question 8
    1 / -0

    If K1 & K2 be first and second ionisation constant of H3PO4 and K1 >> K2 which is incorrect.

    Solution




  • Question 9
    1 / -0

    The degree of hydrolysis of a salt of weak acid and weak base in it's 0.1 M solution is found to be 50%. If the molarity of the solution is 0.2 M, the percentage hydrolysis of the salt should be

    Solution

    % Hydrolysis does notdepend on the conc. in case of “Weak acid + weak base : Salt”

  • Question 10
    1 / -0

    What is the percentage hydrolysis of NaCN in N/80 solution when the dissociation constant for HCN is 1.3 × 10-9 and Kw = 1.0 × 10-14

    Solution

    Weak acid + Strong base : Salt


  • Question 11
    1 / -0

    The compound whose 0.1 M solution is basic is?

    Solution

    Salt of weak acid & strong Base
    CH3COONa → CH3COO– + Na+ CH3COO– + H2O  CH3COOH + OH–
    Basic Solution

  • Question 12
    1 / -0

    Which of the following solution will have pH close to 1.0 ?

    Solution

    75mL of M/5HCl + 25mL of M/5 NaOH 
    Meq of HCl = 15 (∴ Mol. wt. and Eq. wt. of HCl or NaOH are same) 
    Meq of NaOH = 5 Hence, in solution 10meq of HCl remains in 100mL solution. 
    So, concentration of HCl in mixture = 10/100M = M/10 
    In M/10 HCl, [H+] = 10-1 M
    pH = -log[H+] = -log10-1 = 1

  • Question 13
    1 / -0

    The ≈ pH of the neutralisation point of 0.1 N ammonium hydroxide with 0.1 N HCl is

    Solution

    Salt formed : NH4Cl = 0.1 N
    Solution will be slightly acidic due to Hydrolysis

  • Question 14
    1 / -0

    If equilibrium constant of

    CH3COOH + H2O  CH3COO- + H3O+

    Is 1.8 × 10-5, equilibrium constant for

    CH3COOH + OH-  CH3COO- + H2O is

    Solution

    Ka = 1.8 × 10–5
    CH3COO– + H2O  CH3COOH + OH-
    Given reaction is reverse of above

  • Question 15
    1 / -0

    If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [Ka = 2 × 10-4]. The pOH of the resulting solution is

    Solution

    m. equivalent of KOH = 8
    m. equivalent of HCOOH = 16
    Remaining m. eq. (HCOOH) = 8
    Formed m. eq. (HCOOK) = 8
    ⇒ Acidic Buffer
    pH = pKa = 4 – log2
    = 3.7
    pOH = 10.3

  • Question 16
    1 / -0

    A solution with pH 2.0 is more acidic than the one with pH 6.0 by a factor of :

    Solution

    [H+]1 = 10–2 ; [H+]2 = 10–6

  • Question 17
    1 / -0

    The first and second dissociation constants of an acid H2A are 1.0 × 10-5 and 5.0 × 10-10 respectively.

    The overall dissociation constant of the acid will be:

    Solution



  • Question 18
    1 / -0

    An aqueous solution contains 0.01 M RNH2 (Kb= 2 × 10-6) & 10-4 M NaOH.

    The concentration of OH- is nearly:

    Solution



    x2  +  10–4 x – 2 × 10–8 = 0
    x = 10–4
    [OH–] = x + 10–4
    = 2 × 10–4

  • Question 19
    1 / -0

    What volume of 0.2 M NH4Cl solution should be added to 100 ml of 0.1 M NH4OH solution to produce a buffer solution of pH = 8.7 ?            

    Given : pKb of NH4OH = 4.7 ; log 2 = 0.3

    Solution

    pH = 8.7 ⇒ pOH = 5.3
    Basic Buffer

    If volume of salt = V ml

  • Question 20
    1 / -0

    The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be:

    Solution

    Salt of weak acid & weak base

  • Question 21
    1 / -0

    The range of most suitable indicator which should be used for titration of X - Na+ (0.1 M, 10 ml) with 0.1 M HCl should be ( Given : kb(X-) = 10-6 )

    Solution

    NaX + HCl → NaCl + HX
    [HX] at equivalent point = 0.05

    Range of Indicator = 3 to 5

  • Question 22
    1 / -0

    How many gm of solid NaOH must be added to 100 ml of a buffer solution which is 0.1 M each w.r.t. Acid HA and salt Na+ A- to make the pH of solution 5.5. Given pKa(HA) = 5 (Use antilog (0.5)= 3.16)

    Solution



    Suppose x m. mole NaOH was added


  • Question 23
    1 / -0

    The solubility of A2X3 is y mol dm-3. Its solubility product is

    Solution


    Ksp =  (2y)2 (3y)3
    Ksp = 108y5

  • Question 24
    1 / -0

    If Ksp for HgSO4 is 6.4 × 10-5, then solubility of this substance in mole per m3 is

    Solution


    Ksp = S2
    ⇒ 6.4 × 10–5 = S2
    ⇒ S = 8 × 10–3 mole/L
    S = 8 × 10–3 × 103 mole/m3
    ⇒ S = 8 mole/m3

  • Question 25
    1 / -0

    Which of the following in most soluble in water ?

    Solution

    Calculate the solubility ‘s’ for each option, Higher the value of ‘s’ Higher the solubility.

  • Question 26
    1 / -0

    1 M NaCl and 1 M HCl are present in an aqueous solution. The solution is

    Solution

    NaCl + HCl : Not the Buffer and Solution is acidic due to HCl.
    ⇒ pH < 7.

  • Question 27
    1 / -0

    The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized is:

    Solution


    50% ionised ⇒ [Salt] = [Acid]
    ⇒ pH = pKa = 4.5
    ⇒ pOH = 9.5

  • Question 28
    1 / -0

    The precipitate of CaF2 (Ksp = 1.7 × 10-10) is obtained when equal volumes of the following are mixed

    Solution

    For ppt Qsp > Ksp
    CaF2→ Ca2++2F–
    Qsp  = (Ca2+) (F–)2
    (A) Qsp  = 12.5 × 10–14
    (B) Qsp  = 12.5 × 10–10
    (C) Qsp  = 12.5 × 10–13
    (D) Qsp  = 12.5 × 10–15
    Only (B) option will get precipitate.

  • Question 29
    1 / -0

    pH of saturated solution of silver salt of monobasic acid HA is found to be 9.

    Find the Ksp of sparingly soluble salt Ag A(s).

    Given : Ka(HA) = 10-10

    Solution




    Ksp  = S (S–X) = 11 × 10–6 × 10–6
    = 1.1 × 10–11

  • Question 30
    1 / -0

    When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8 × 10-10) will occur only with:

    Solution

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