Ncert Solutions Chemistry Class 11th
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New answer posted
10 months agoContributor-Level 10
The hydrogen ion concentration in the given substances can be calculated by using the given relation: pH=−log [H+]
Hence, [H+] = 10−pH
Milk: [H+] = 10−6.8 = 1.58*10−7M
Black coffee: [H+] = 10−5.0 =1*10−5M
Tomato juice: [H+] = 10−4.2 =6.31*10−5M
Lemon juice: [H+]=10−2.2 = 6.31*10−3M
Egg white: [H+]=10−7.8=1.58*10−8M
New answer posted
10 months agoContributor-Level 10
a. Human muscle fluid 6.83
pH=6.83
pH=−log [H+]
∴6.83=−log [H+]
[H+]= 1.48 x 10−7M
b. Human stomach fluid, 1.2:
pH=1.2
1.2=−log [H+]
∴ [H+] = 0.063 M = 6.3 x 10-2 M
c. Human blood, 7.38:
pH=7.38=−log [H+]
∴ [H+]= 4.17 x 10−8M
d. Human saliva, 6.4:
pH=6.4
6.4=−log [H+]
[H+]= 3.98 x 10−7 M
New answer posted
10 months agoContributor-Level 10
Kb= 5.4*10−4
c= 0.02M
Then, α= (Kb /c)1/2
α= (5.4*10−4 / 2 x 10-2)1/2 =0.1643
(CH3)2NH+H2O ↔ (CH3)2NH+2+OH-
[ (CH3)2NH] = 0.02 – x ≈ 0.02
[ (CH3)2NH+2] = x
[OH-] = 0.1 + x
≈ 0.1
Now, Kb= [ (CH3)2NH+2] [OH−]/ [ (CH3)2NH] = (x * 0.1) / (0.025).
x = 1.08 x 10-4
% of dimethylamine ionised = (1.08 x 10-4) x (100 / 0.02) = 0.54%
New answer posted
10 months agoContributor-Level 10
pKa? =? logKa= 4.74
Ka? = 10? pKa =10?4.74 = 1.8*10?5
Let x be the degree of dissociation. The concentration of acetic acid solution, C = 0.05 M
The degree of dissociation,
x= (Ka / C)1/2? = (1.8*10?5 / 0.05)1/2 ? = 0.019
(a) The solution is also 0.01 M in HCl.
Let x M be the hydrogen ion concentration from ionization of acetic acid. The hydrogen ion concentration from ionization of HCl is 0.01 M. The total hydrogen ion concentration
[H+] = 0.01 + x
The acetate ion concentration is equal to the hydrogen ion concentration from ionization of acetic acid. This is also equal to the concentration of acetic acid that has dissociated.
[CH3?
New answer posted
10 months agoContributor-Level 10
Let c be the initial concentration of C6H5NH3+ and x be the degree of ionisation.
C6H5NH2 + H2O? C6H5NH3+ + OH-
c (1-x) cx cx
Kb = [C6H5NH3+] [ OH-] / [C6H5NH2]
= [cx] [cx] / [c (1 – x)]
Since x is very small and negligible 1 – x≈ 1
∴Kb= [cx] [cx] / [c] = cx2
=> x =
=
= 6.56 x 10-4
∴ [OH-] = cx = 0.001 x 6.56 x 10-4 = 6.56 x 10-7 M
[H+]= Kw / [OH-] = 10-14 / 6.56 x 10-7 = 1.52 x 10-8
pH= –log [H+] = –log1.52 x 1
New answer posted
10 months agoContributor-Level 10
pH = 9.95,
pOH = 14 – pH = 14 − 9.95 = 4.05
[OH−] = 10−pOH = 10−4.05 = 8.913 * 10−5
Codeine + H2? O? CodeineH+ + OH−
The ionization constant, Kb? = [CodeineH+] [OH−] / [codeine]?
= [ (8.913*10−5)* (8.913*10−5)] / 5*10−3
= 1.588*10−6.
pKb? = −log (1.588*10−6)
= 5.8
New answer posted
10 months agoContributor-Level 10
H+] = cα = 0.1 * 0.132 = 0.0132M
pH = −log [H+] = −log0.0132
= 1.88
The acid dissociation constant is
Ka? = cα2? / (1−α) = 0.1 * (0.132)2 / (1−0.132)
= 2.01*10−3.
pKa? = −logKa? = −log (2.01*10−3) ≈ 2.7
New answer posted
10 months agoContributor-Level 10
(a) For 2g of TlOH dissolved in water to give 2 L of solution:
[TlOH] = [OH−] = (2*1)? / (2*221) = (1 / 221)? M
pOH = −log [OH]− = −log (1/221)?
= 2.35
pH = 14 – pOH = 14 − 2.35 = 11.65
(b) For 0.3 g of Ca (OH)2? dissolved in water to give 500 mL of solution:
[OH−] = 2 [Ca (OH)2? ] = 2 (0.3*1000/500? ) = 1.2M
pOH = −log [OH−] = −log1.2 = 1.79
pH= 14−pOH=14−1.79
=12.21
(c) For 0.3 g of NaOH dissolved in water to give 200 mL of solution:
[OH−]= [NaOH] = 0.3*1000/200? = 1.5M
pOH= −log [OH−] = −log1.5 = 1.43
pH= 14 – pOH = 14 − 1.43
= 12.57
(d) For 1mL of 13.6 M HCl diluted w
New answer posted
10 months agoContributor-Level 10
(a) 0.003 M HCl
[H3? O+] = [HCl] = 0.003M
pH = −log [H+] = −log (3.0*10−3) = 2.523
(b) 0.005 M NaOH
[OH−] = [NaOH] = 0.005M
[H+] = Kw? / [OH−]? = 10−14/ 0.005? =2*10−12
pH= −log [H+]=−log (2*10−12)=11.699
(c) 0.002M HBr
[H+]= [HBr]=0.002
pH= −log [H+]=−log0.002=2.699
(d) 0.002M KOH
[OH−]= [KOH]=0.002M
[H+]= Kw / [OH−]? =10−14 / 0.002? =5*10−12
pH= −log [H+]=−log (5*10−12)=11.301
New answer posted
10 months agoContributor-Level 10
pH= −log [H+]=4.15
[H+]= antilog (−4.15)= 7.08*10−5
[A−]= [H+]=7.08*10−5
The concentration of undissociated acid is 0.01−0.000071=0.009929M.
HA+H2? O? H3? O++A−
Ka? = [H3? O+] [A−] / [HA]? = (7.08*10−5) (7.08*10−5)? / 0.009929
= 5.05*10−7
pKa? = −logKa? = −log5.05*10−7 ≈ 6.3
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