The Total solution for NCERT class 6-12
The pH of 0.005M codeine (C18H21NO3) solution is 9.95.Calculate its ionization constant and pKb.
Answer - 51 : -
c = 0.005
pH= 9.95
pOH= 4.05
pH= – log (4.105)
Whatis the pH of 0.001 M aniline solution? The ionization constant of aniline canbe taken from Table 7.7. Calculate the degree of ionization of aniline in thesolution. Also calculate the ionization constant of the conjugate acid ofaniline.
Answer - 52 : -
Kb = 4.27 × 10–10
c = 0.001M
pH=?
α=?
Thus,the ionization constant of the conjugate acid of aniline is 2.34 × 10–5.
Calculate the degree of ionization of 0.05Macetic acid if its pKa value is 4.74.
Howis the degree of dissociation affected when its solution also contains (a) 0.01M (b) 0.1 M in HCl?
Answer - 53 : -
When HCl is added to the solution, theconcentration of H+ ions will increase. Therefore, theequilibrium will shift in the backward direction i.e., dissociation of aceticacid will decrease.
Case I: When 0.01 M HCl is taken.
Let x be the amount ofacetic acid dissociated after the addition of HCl.
As the dissociation of a very small amountof acetic acid will take place, the values i.e., 0.05 – x and0.01 + x can be taken as 0.05 and 0.01 respectively.
Case II: When 0.1 M HCl is taken.
Let the amount of acetic acid dissociated inthis case be X. As we have done in the first case, theconcentrations of various species involved in the reaction are:
The ionization constant of dimethylamine is5.4 × 10–4. Calculate its degree of ionization in its0.02 M solution. What percentage of dimethylamine is ionized if the solution isalso 0.1 M in NaOH?
Answer - 54 : -
Now,if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base)undergoes complete ionization.
And,
Itmeans that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will getdissociated.
Calculatethe hydrogen ion concentration in the following biological fluids whose pH aregiven below:
(a)Human muscle-fluid, 6.83
(b)Human stomach fluid, 1.2
(c)Human blood, 7.38
(d)Human saliva, 6.4.
Answer - 55 : -
(a) Human musclefluid 6.83:
pH= 6.83
pH = – log [H+]
∴6.83 = – log [H+]
[H+] =1.48 × 10–7 M
(b) Human stomachfluid, 1.2:
pH=1.2
1.2 = – log [H+]
∴[H+] = 0.063
(c) Human blood,7.38:
pH = 7.38 = – log [H+]
∴ [H+] = 4.17 × 10–8 M
(d) Human saliva,6.4:
pH= 6.4
6.4 = – log [H+]
[H+] = 3.98 × 10–7
ThepH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0,4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ionconcentration in each.
Answer - 56 : -
Thehydrogen ion concentration in the given substances can be calculated by usingthe given relation:
pH = –log [H+]
(i)pH of milk = 6.8
Since, pH = –log [H+]
6.8 = –log [H+]
log [H+] = –6.8
[H+] = anitlog(–6.8)
(ii)pH of black coffee = 5.0
5.0 = –log [H+]
log [H+] = –5.0
[H+] = anitlog(–5.0)
(iii)pH of tomato juice = 4.2
4.2 = –log [H+]
log [H+] = –4.2
[H+] = anitlog(–4.2)
(iv)pH of lemon juice = 2.2
2.2 = –log [H+]
log [H+] = –2.2
[H+] = anitlog(–2.2)
(v)pH of egg white = 7.8
7.8 = –log [H+]
log [H+] = –7.8
[H+] = anitlog(–7.8)
If0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K.Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What isits pH?
Answer - 57 : -
The solubility of Sr(OH)2 at 298 K is 19.23g/L of solution. Calculate the concentrations of strontium and hydroxyl ionsand the pH of the solution.
Answer - 58 : -
Solubility of Sr(OH)2 = 19.23 g/L
Then, concentration of Sr(OH)2
The ionization constant of propanoic acid is1.32 × 10–5. Calculate the degree of ionization of theacid in its 0.05M solution and also its pH. What will be its degree ofionization if the solution is 0.01M in HCl also?
Answer - 59 : -
Letthe degree of ionization of propanoic acid be α.
Then,representing propionic acid as HA, we have:
Inthe presence of 0.1M of HCl, let α´ be the degree of ionization.
ThepH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionizationconstant of the acid and its degree of ionization in the solution.
Answer - 60 : -
c = 0.1 M
pH= 2.34