Question

[A] - Pa(KH) H=0.033 TAJ = 0.033 12.0 atm) =10.066 M 3. Calculate the concentration of TIH(aq) in the three scenarios below: TICI(s) = TI+(aq) + Cl-(aq) Ksp = 1.4 x 10-4 a. You add 1.4 x 10-4 moles TICI to 1 liter of water. Ksp = [Tit][Cl] 1.4*10-4 (1.4×109) = x² TiCIJ 11.4 x 10-4 M = x 1.48104 b. You add 1.2 x 10-2 moles TICI to 1 liter of water. c. You add 1.5 moles TICI to 1 liter of water.
just B and C, Thanks!

Answer 1

b)

Balanced chemical equation:-

TlCl(s) <-----> Tl+(aq) + Cl-(aq)

It is evident from the balanced chemical equation that 1 mole of TlCl produces 1 mole of Tl+ and 1 mole of Cl-.

i.e., x moles of TlCl will produce x moles of Tl+ and x mole of Cl-.

Filling Ice table

	TlCl	Tl+	Cl-
I	0.012	0	0
C	-x	+x	+x
E	0.012-x	x	x

Here,

I stands for Initial no. of moles.

C stands for change in no. of moles, -ve sign denotes loss and +ve sign denotes gain.

E stands for equilibrium/final no. of moles.

Finding concentration using the formula-

$Concentration (M) = \tfrac{No. of moles}{Volume in litre} ------(a)$

Volume of water = 1 L -----(Given)

Concentration of Tl+ = (x/1) ------using equation(a)

[Tl+] = x M

Concentration of Cl- = (x/1) ------using equation(a)

[Cl-] = x M

Expression for Ksp can be written as-

$K_{sp} = \tfrac{[Tl^{+}][Cl^{-}]}{[TlCl]} ------(b)$

Here,

Ksp = The solubility product constant

[Tl+] = Concentration of Tl+

[Cl-] = Concentration of Cl-

[TlCl] = Concentration of TlCl

In the given chemical reaction, TlCl is solid, the concentration of solid in expression for Ksp is taken as 1.

Therefore, [TlCl] = 1

Hence, equation(b) can be written as-

Ksp = [Tl+][Cl-] ------(b')

Ksp = 1.4X10^-4 ------(Given)

= 0.00014

Putting all values in equation(b'), we get,

0.00014 = x.x

0.00014 = x^2

x = $\sqrt{0.00014}$

x = 0.0118 M

[Tl+] = 0.0118 M

The concentration of Tl+ in the given scenario is 0.0118 M.

c)

Balanced chemical equation:-

TlCl(s) <-----> Tl+(aq) + Cl-(aq)

It is evident from the balanced chemical equation that 1 mole of TlCl produces 1 mole of Tl+ and 1 mole of Cl-.

i.e., x moles of TlCl will produce x moles of Tl+ and x mole of Cl-.

Filling Ice table

	TlCl	Tl+	Cl-
I	1.5	0	0
C	-x	+x	+x
E	1.5-x	x	x

Here,

I stands for Initial no. of moles.

C stands for change in no. of moles, -ve sign denotes loss and +ve sign denotes gain.

E stands for equilibrium/final no. of moles.

Finding concentration using the formula-

$Concentration (M) = \tfrac{No. of moles}{Volume in litre} ------(a)$

Volume of water = 1 L -----(Given)

Concentration of Tl+ = (x/1) ------using equation(a)

[Tl+] = x M

Concentration of Cl- = (x/1) ------using equation(a)

[Cl-] = x M

Expression for Ksp can be written as-

$K_{sp} = \tfrac{[Tl^{+}][Cl^{-}]}{[TlCl]} ------(b)$

Here,

Ksp = The solubility product constant

[Tl+] = Concentration of Tl+

[Cl-] = Concentration of Cl-

[TlCl] = Concentration of TlCl

In the given chemical reaction, TlCl is solid, the concentration of solid in expression for Ksp is taken as 1.

Therefore, [TlCl] = 1

Hence, equation(b) can be written as-

Ksp = [Tl+][Cl-] ------(b')

Ksp = 1.4X10^-4 ------(Given)

= 0.00014

Putting all values in equation(b'), we get,

0.00014 = x.x

0.00014 = x^2

x = $\sqrt{0.00014}$

x = 0.0118 M

[Tl+] = 0.0118 M

The concentration of Tl+ in the given scenario is 0.0118 M.