The conjugate acid of NH2- is
Conjugate acid of a base is formed by adding H+ to it
NH2- + H+ → NH3
Out of the following, amphiprotic species are
I : HPO32-
Amphiprotic : can accept and Release H+
Only H2PO–4 & HCO–3
pH of an aqueous solution of NaCl at 85°C should be
NaCl Solution : pH is the, pH of water.
pH at 25°C <7
1 c.c. of 0.1N HCl is added to 99 CC solution of NaCl. The pH of the resulting solution will be
Volume of resulting solution = 100 ml
10 ml of is mixed with 40 ml of . The pH of the resulting solution is
[H+] = 10–2
pH = 2.
If pKb for fluoride ion at 25°C is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is :
H+ H+ + F–
pKw = pKa + pKb
[For conjugate Acid-Base]
⇒ pKa = 14 – 10.87 = 3.17
Ka = 6.76 × 10–4
The pH of an aqueous solution of 1.0 M solution of a weak monoprotic acid which is 1% ionised is:
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
If K1 & K2 be first and second ionisation constant of H3PO4 and K1 >> K2 which is incorrect.
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
% Hydrolysis does notdepend on the conc. in case of “Weak acid + weak base : Salt”
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
Weak acid + Strong base : Salt
The compound whose 0.1 M solution is basic is?
Salt of weak acid & strong Base
CH3COONa → CH3COO– + Na+ CH3COO– + H2O CH3COOH + OH–
Which of the following solution will have pH close to 1.0 ?
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
The ≈ pH of the neutralisation point of 0.1 N ammonium hydroxide with 0.1 N HCl is
Salt formed : NH4Cl = 0.1 N
Solution will be slightly acidic due to Hydrolysis
If equilibrium constant of
CH3COOH + H2O CH3COO- + H3O+
Is 1.8 × 10-5, equilibrium constant for
CH3COOH + OH- CH3COO- + H2O is
Ka = 1.8 × 10–5
CH3COO– + H2O CH3COOH + OH-
Given reaction is reverse of above
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
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
pOH = 10.3
A solution with pH 2.0 is more acidic than the one with pH 6.0 by a factor of :
[H+]1 = 10–2 ; [H+]2 = 10–6
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:
An aqueous solution contains 0.01 M RNH2 (Kb= 2 × 10-6) & 10-4 M NaOH.
The concentration of OH- is nearly:
x2 + 10–4 x – 2 × 10–8 = 0
x = 10–4
[OH–] = x + 10–4
= 2 × 10–4
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
pH = 8.7 ⇒ pOH = 5.3
If volume of salt = V ml
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:
Salt of weak acid & weak base
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 )
NaX + HCl → NaCl + HX
[HX] at equivalent point = 0.05
Range of Indicator = 3 to 5
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)
Suppose x m. mole NaOH was added
The solubility of A2X3 is y mol dm-3. Its solubility product is
Ksp = (2y)2 (3y)3
Ksp = 108y5
If Ksp for HgSO4 is 6.4 × 10-5, then solubility of this substance in mole per m3 is
Ksp = S2
⇒ 6.4 × 10–5 = S2
⇒ S = 8 × 10–3 mole/L
S = 8 × 10–3 × 103 mole/m3
⇒ S = 8 mole/m3
Which of the following in most soluble in water ?
Calculate the solubility ‘s’ for each option, Higher the value of ‘s’ Higher the solubility.
1 M NaCl and 1 M HCl are present in an aqueous solution. The solution is
NaCl + HCl : Not the Buffer and Solution is acidic due to HCl.
⇒ pH < 7.
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:
50% ionised ⇒ [Salt] = [Acid]
⇒ pH = pKa = 4.5
⇒ pOH = 9.5
The precipitate of CaF2 (Ksp = 1.7 × 10-10) is obtained when equal volumes of the following are mixed
For ppt Qsp > Ksp
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.
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
Ksp = S (S–X) = 11 × 10–6 × 10–6
= 1.1 × 10–11
When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8 × 10-10) will occur only with: