Class 11th
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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
New answer posted
6 months agoContributor-Level 10
c=0.1M
pH= −log [H+]
=> 2.34 = −log [H+]
So, −log [H+]= 2.34
=> [H+]= 4.5*10−3
Also,
[H+]= cα
=>4.5*10−3= 0.1*α
=> α=4.5*10−2= 0.045
Then,
Ka = α2/c = (45*10−3)2/ 0.1
=202.5*10−6
=2.02*10−4
New answer posted
6 months agoContributor-Level 10
Let the degree of ionization of propanoic acid be α. Then, representing propionic acid as HA, we have:
HA + H2O ⇔ H3O+ + A−
(.05−0.0α)≈0.5 0.05α 0.05α
Ka= [H3O+] [A−] / [HA]
= (0.05α) (0.05α) / 0.05
= 0.05α2
=> α = (Ka /0.05)1/2
&nbs
New answer posted
6 months agoContributor-Level 10
Solubility of Sr (OH)2=19.23g/L
The molecular weight of Sr (OH)2 is 87.6 + 2 (17)=121.6
Then, concentration of Sr (OH)2=19.23 /121.63M=0.1581M
Sr (OH)2 (aq)→Sr2+ (aq)+2 (OH−) (aq)
∴ [Sr2+]=0.1581M
[OH−]=2*0.1581M=0.3126M
Now, Kw= [OH−] [H+]
=> [H+] = 10−14 / 0.3126
=> [H+]=3.2*10−14
∴pH= 13.495
New answer posted
6 months agoContributor-Level 10
[KOH]= [K+]= [OH−]= (0.561*1000) / (56*200)? =0.050M
[H+]=Kw / [OH−]? =10−14 / 0.05? =2.0*10−13
pH=−log [H+]=−log (2.0*10−13)
=12.7
New answer posted
6 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
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