Ncert Solutions Chemistry Class 11th
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New answer posted
6 months agoContributor-Level 10
Since pH=3.19,
[H3O+]=6.46*10−4M
C6H5COOH+H2O↔C6H5COO−+H3O
Ka= [C6H5COO−] [H3O+] / [C6H5COOH]
[C6H5COOH] / [C6H5COO−]= [H3O+] / Ka=6.46*10−4 / 6.46*10−5=10
Let the solubility of C6H5COOAg be xmol/L.
Then,
[Ag+]=x
[C6H5COOH]+ [C6H5COO−]=x
10 [C6H5COO−]+ [C6H5COO−]=x
[C6H5COO−]=x / 11
Ksp = [Ag+] [C6H5COO−]
= >2.5*10−13=x (x / 11)
= >x=1.66*10−6mol/L
Thus, the solubility of silver benzoate in a pH3.19 solution is 1.66*10−6mol/L.
Now, let the solubility of C6H5COOAg be x′mol/L.
Then, [Ag+]=x′Mand [C6H5COO−]=x′M
Ksp= [Ag+] [C6H5COO−]
Ksp= (x′)2
x′= (Ksp)1/2= (2
New answer posted
6 months agoContributor-Level 10
After mixing, the concentration of NaIO3? is 0.002 / 2? =0.001M.
After mixing, the concentration of Cu (ClO3? )2? is 0.002? / 2 =0.001M.
NaIO3? Na++IO3−?
[IO3−? ]=0.001M
Cu (ClO3? )2? Cu2++2ClO3−?
[Cu2+]=0.001M
The ionic product of Cu (IO3? )2? is
[Cu2+] [I−]2=0.001*0.0012=1*10−9
As the ionic product is less than the solubility product, no precipitation will occur.
New answer posted
6 months agoContributor-Level 10
Silver chromate: Ag2CrO4? ? 2Ag+ + CrO42−?
[Ag+] = 2s1? , CrO42−? = s1?
Ksp? = (2s1? )2 (s1? ) = 4s3 = 1.1*10−12
s1? = 6.5 * 10−5 . (1)
Silver bromide: AgBr? Ag+ + Br−
[Ag+] = [Br−] = s2?
Ksp? = (s2? ) * (s2? ) = (s2)2? = 5.0 * 10−13
s2? =7.07*10−7. (2)
Divide equation (1) by equation (2) to obtain the ratio of the molarities of saturated solutions:
? s1/s2? = (6.50*10−5)/ (7.07*10−7)? = 91.9
New answer posted
6 months agoContributor-Level 10
1. Silver chromate:
Ag2CrO4→2Ag++CrO42−
Then, [Ag+] = (2s), [CrO42−] = s
Ksp= [Ag+]2 [CrO42−]
s = 0.65*10−4M
So, [Ag+] = 2s = 1.30 x 10−4M, [CrO42−] = 6.5 x 10-5M
2. Barium Chromate:
BaCrO4→Ba2++CrO42−
[Ba2+] = [CrO42−] = s
Then, Ksp= [Ba2+] [CrO42−] = s x s
= > 1.2 x 10-10M = s2
= > s = 1.09 x 10-5M
3. Ferric Hydroxide:
Fe (OH)3→Fe3+ + 3OH−
Then [Fe3+] = s, [OH−] = 3s
Ksp= [Fe3+] [OH−]3
Let s be the solubility of Fe (OH)3
[Fe3+] = s = 1.38 x 10-10M
[OH−] = 3s = 4.14 x 10-10M
4. Lead Chloride:
PbCl2→Pb2++2Cl−
Then, [Pb2+] = s, [Cl−] = 2s
Ksp= [Pb2+] [Cl−]2
= >Ksp=s x (2s)2 =4s3
⇒1.6*10−5=4s3
⇒0.4*10
New answer posted
6 months agoContributor-Level 10
(a) Total number of moles present in 10 mL of 0.2 M calcium hydroxide are (10*0.2) / 1000? = 0.002 moles.
Total number of moles present in 25 mL of 0.1 M HCl are (25*2) / 1000? = 0.0025 moles.
Ca (OH)2? + 2HCl → CaCl2? + 2H2? O
1 mole of calcium hydroxide reacts with 2 moles of HCl.
0.0025 moles of HCl will react with 0.00125 moles of calcium hydroxide.
Total number of moles of calcium hydroxide unreacted are 0.002−0.00125 = 0.00075 moles.
Total volume of the solution is 10 + 25 = 35 mL.
The molarity of the solution is (0.00075*1000) / 35? = 0.0214M
[OH−] = 2 * 0.0214 = 0.0428M
pOH = −log0.0428 = 1.368
New answer posted
6 months agoContributor-Level 10
Ionic product,
Kw= [H+] [OH−]
Let [H+]= x
Since [H+]= [OH−], Kw=x2.
⇒Kw at 310K is 2.7*10−14
∴2.7*10−14=x2
⇒x=1.64*10−7
⇒ [H+]=1.64*10−7
⇒ pH= −log [H+]=−log [1.64*10−7]=6.78
Hence, the pH of neutral water is 6.78.
New answer posted
6 months agoContributor-Level 10
Given Ka=1.35 x 10−3
For acid solution:
[H+] = (KaC)1/2 = (0.00135 x 0.1)1/2 = 0.0116M
pH = – log [H+] = –log0.0116 = 1.936
For 0.1M sodium salt solution
ClCH2COONa is the salt of a weak acid i.e., ClCH2COOH and a strong base i.e., NaOH.
pKa = -logKa = log (0.00135) = 2.8697
pKw = 14
logc = log0.1 = −1
pH = 0.5 [pKw + pKa + logc] = 0.5 [14 + 2.8697 -1]
= 7.935
New answer posted
6 months agoContributor-Level 10
(i) NaCl:
NaCl+H2O↔NaOH+HCl
Strong base Strong acid
Therefore, it is a neutral solution.
(ii) KBr:
KBr+H2O↔KOH+HBr
Strong base Strong acid
Therefore, it is a neutral solution.
(iii) NaCN:
NaCN+H2O↔HCN+NaOH
Weak acid Strong base
Therefore, it is a basic solution.
(iv) NH4NO3
NH4NO3+H2O↔NH4OH+HNO3
Weak base Strong acid
Therefore, it is an acidic solution.
(v) NaNO2
NaNO2+H2O↔NaOH+HNO2
Strong base Weak acid
Therefore, it is a basic solution.
(vi) KF
KF+H2O↔KOH+HF
Strong base Weak acid
Therefore, it is a basic solution.
New answer posted
6 months agoContributor-Level 10
pH=3.44
We know that,
pH=−log [H+]
∴ [H+]=3.63*10−4
Then, Kb= (3.63*10−4)2 / 0.02 (? concentration =0.02M)
⇒Kb=6.6*10−6
Now, Kb=Kw / Ka
⇒Ka=Kw / Kb=10−14 / 6.6*10−6=1.51*10−9
New answer posted
6 months agoContributor-Level 10
Since NaNO2 is the salt of a strong base (NaOH) and a weak acid (HNO2);
NO2− + H2O ↔ HNO2 + OH−
Then, Kb= [HNO2] [OH−] / [NO2−]
⇒ Kw/ Ka = (10−14) / (4.5*10−4)
= 0.22*10−10
Now, if x moles of the salt undergo hydrolysis, and then the concentration of various species present in the solution will be:
[NO2−]= 0.04−x; 0.04
[HNO2]= x
[OH−]= x
Kb= x2 / 0.04
= 0.22*10−10
x2= 0.0088*10−10
x= 0.093*10−5
∴ [OH−]= 0.093*10−5 M
[H3O+]=10−14 / 0.093*10−5=10.75*10−9 M
⇒ pH= −log (10.75*10−9)=7.96
Now, degree of hydrolysis
= x / 0.04= (0.093*10−5)/ 0.04
= 2.325*10−5
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