Equilibrium
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New answer posted
4 months agoContributor-Level 10
CaSO4 (s)↔Ca2+ (aq)+SO2−4 (aq)
Ksp= [Ca2+] [ SO2−4]
Let the solubility of CaSO4 be s.
[Ca2+] = [ SO2−4] = s
Then, Ksp=s2
9.1*10−6=s2
s =3.02*10−3mol/L
Molecular mass of CaSO4=136g/mol
Solubility of CaSO4 in gram/L= 3.02*10−3*136=0.41g/L
This means that we need 1L of water to dissolve 0.41g of CaSO4.
Therefore, to dissolve 1g of CaSO4 we require = 10.41L= 2.44Lof water.
New answer posted
4 months agoContributor-Level 10
Let the maximum concentration of each solution be x mol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., x/2.
∴ [FeSO4]= [Na2S]=x / 2
Then, [Fe2+]= [FeSO4]=x/ 2
Also, [S2−]= [Na2S]=x/2
FeS (x)↔Fe2+ (aq)+S2− (aq)
Ksp= [Fe2+] [S2−]
= >6.3*10−18= (x/2) (x/2)
x2/4=6.3*10−18
⇒x= 5.02*10−9
If the concentrations of both solutions are equal to or less than 5.02*10−9M, then there will be no precipitation of iron sulphide.
New answer posted
4 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
4 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
4 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
4 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
4 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
4 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
4 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
4 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.
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