Here , [H+] for acid is < 10 ^-7 { which is not possible }

so,

total [H+] = 10 ^-7 + 10 ^-9

= 1.01 x 10 ^-7

Thus, pH = - log [H+]

= 6.99

Limitations of Ostwald's Dilution law are :

• Applicable only for weak electrolytes
• Applicable only for highly diluted solutions
• Assumption is made for 1- I_cto be equal to 1 being I_cvery low for making calculations simpler but this assumption can alter the exact data.

Bronsted bases are the only bases which are capable to give hydroxyl ion in solution which are also explained by lewis, but lewis also explains the basic nature of substances which doesn't yield hydroxyl ion and doesn't also require solution like NH₃, CH3^- etc. Hence, All species that has tendency to donate non-bounded electrons including OH- are lewis bases but may or may not be Bronsted bases. So, the statement is true.

A Lewis acid is any substance, such as the H⁺ion, that can accept a pair of nonbonding electrons.

A Lewis base isany substance, such as the OH^-ion, that can donate a pair of nonbonding electrons.

Electron pair acceptor are acid and donating are bases.

according to Bronsted and Lowry,

Any chemical species capable of donating a proton [H+] when dissolved in water is an acid.

Any chemical species capable of accepting proton [H+], which has a lone pair of electrons to make a bond with [H+] is a base.

• The solubility productconstant istheequilibrium constant forthedissolutionof asolid substance into an aqueous solution. It is denoted bythesymbol Ksp.The solubility productisakindofequilibrium constant and its value depends on temperature. Ksp usually increases with an increase in temperature due to increasedsolubility.
• 
  

the mathematical expression representing the variation of the degree of dissociation with dilution for weak electrolytes is called OStwald's dilution law.

mathematical exp, K = I_d ²/ (1-I_d ) V

• The main difference between strong electrolytes and weak electrolytes is thatstrong electrolytes can almost completely dissociate into its ionswhereas weak electrolytes partially dissociate into ions.

Since, according to the question;

200 ml of an aqueous solution of HCl is mixed with 300 ml of an aqueous solution of NaOH.

The given pH of HCl is 2;

The given pH of NaOH is 12;

We have to find the resultant pH.

[HCl] =

[NaOH] =

So, the reaction goes like this :

Since , before reaction;

After reaction;

0 1 2 2

We will be finding the concentration of the hudroxide ion first.

left from NaOH =M

The Ph of the hydroxide ion is calculated as

Since , we know that

Theof the hydrogen ion...

It is a measure for strength of electrolytes, the electrolytes which have degree of ionization nearly equal to 1 are strong, else are weak.

It is the representation of ratio of molecules disassociated ions to the total number of molecules when put in water. its range is [0,1].

Number of moles of NaF in 25.0 cm^3

0.02 /1000 * 25.0 = 5.0 x 10^-4

The total volume of the solution on mixing Ba(NO3)2 and NaF solutions

= 25.0 + 25.0

= 50.0cm^3

This means that 1.25 10^-3 moles of Ba(NO3) and 5.0 10^-4 moles of NaF are now present in 50cm3

.: The concentration of Ba(NO3)2 in solution

(1.25 10^-3 )/50 x 1000

= 0.025 mol L^-1

and, the concentration of NaF in solutions

= (5.0x 10^-4)/50x 1000

= 0.01 mol L^-1

Hence, | Ba?+| = | Ba(NO3)2] in solution

= 0.025 x (0.01)^2

= 2.5 x 10^-6

The...

Let the solubility of$Ca(OH{)}_2$in pure water
= S moles/litre
$Ca(OH{)}_{2}C{a}^{2+}+2O{H}^{}$
S 2S
$Then{K}_{sp}=\left[C{a}^{2+}\right]\left[O{H}^{}\right]$
$4.42{10}^5=S(2S{)}^2$
$4.42{10}^5=4S^3$
$S=2.224{10}^2=0.0223moleslitr{e}^1$
No. of moles of$C{a}^{2+}$ions in 500 ml of solutions = 0.011. Now when 500 ml of saturated solution is mixed with 500 ml of 0.4 M NaOH, the resultant volume is 1000 ml. The molarity of$O{H}^{}$ions on the resultant solution would therefore be 0.2 M
Maximum$\left[C{a}^{2+}\right]$that can be tolerated$=\frac{ksp}{{\left[O{H}^{}\right]}^2}$
$=\frac{4.42{10}^5}{(0.2{)}^2}=0.0011M$
Thus number of moles of$C{a}^{2+}$or$Ca(OH{)}_2$precipitated...

NaOH ------> Na+ + OH-

given, [NaOH] = 0.4 g/liter

molecular mass of NaOH is 40 g/mol.

thus, Molarity of the solution = 0.4 / 40

=1/100 = 10^-2mole/liter

now, from the dissociation pattern of NaOH

[NaOH] = [OH-] = 10^-2

thus, pOH = -log [OH-]

= 2

thus, pH = 14 - pOH

= 12

Given,

pH of NaOH is 11

i.e. - log [ H+ ] = 10^-11

thus, [OH-] = 10 ^-14 / 10 ^-11

= 10 ^-3 mole/liter

molecular weight of NaOH is 40 g/mole

thus, in the strength is g/liter = 40 x 10 ^-3

= 0.04 g/liter

given,

pH = 10

i.e. - log [ H⁺] = 10

thus, [H⁺] = 1 x 10 ^-10

we know, [H⁺] [OH^-] = 10 ^-14

SO, [OH-] = 1 X 10 ^-4 mole/litre

CaF2 -----------> Ca⁺⁺ + F2-

[ Ca++ ] = 2.94 x 10 ^-4

2 [ F- ] = 2 x 2.92 x 10 ^-4

thus , Ksp = [ Ca⁺⁺] [ 2F^-]²

= 4 x 2.94 x 10 ^-4 x ( 2.94 x 10 ^-4) ²

= 1.01 x 10 ^-10

For weakly soluble electrolytes which gives less ions when dissolved in water the solubility product is the product of concentrations of ions expressed in terms of molarity produced during the ionization of the electrolyte.

its value is very less ( < 10^-5).

HNO₃ ------------> H⁺ + NO₃^-

given,

concentration of acid = 0.04 N = 0.04 M

thus, [H⁺] = 0.04 M

pH = - log [ 0.04 ]

= - log [ 0.04 ]

= 1.39

HCN being a weak acid has low ionization constant,

C = 0.1 M

pH = 5.2

thus [H⁺] = 10 ^{- 5.2} = 0.1 I_c

I_c= 6.3 x 10 ^-5

now, K_a = Ic ² x C [ for weak acid ]

=( 6.3 x 10 ^-5 )² x 0.1

= 3.98 x 10 ^-10

given that , 30 ml of N/10 NaOH was required ,
Applying Normality equation,
N1V1 = N2V2
N1V1 = ( 30 / 1000 ) x (1/10) ------1
Again, 10 ml of diluted acid acid was taken
N1 = 30 / 10 x (1/10)
N1 = 0.3
This means the normality of diluted acid was 0.3N
Since 10 ml of the solution was diluted ,
Mass of acid in 10 ml ,
mass = NEVml /1000
=0.3 x ( 98/2 ) x 10 / (1000)
= 0.147g
Since the total solution was of 1 litre,
Total mass of acid = ( 0.147 / 10 ) x 1000
= 14.7 g
Percentage purity...

Chemicalequilibriumrefers to thestateof a systeminwhich the concentration of the reactant and the concentration of the products do not change with time and the system does not display any further changeinproperties.