To find out the concentration, we apply the formula:

Concentration=

Where n= No. of moles, V= Volume

As we know that n= ()

∴ Concentration=

To calculate [H⁺], we apply the formula:

K_w = [H⁺] [OH^-]

Where K_w is the ionic product of water which is defined as the product of molar concentration of H⁺ and OH^- ions.

To calculate pH, we apply the formula:

pH = -log[H⁺]

a) For 2g of TiOH dissolved in water to give 2L of solution,

Given:

Mass of TiOH(m) = 2g

Volume(V) = 2l

Molar mass of TiOH(M) = 221 g/mol

To calculate concentration of TiOH, we apply the formula,

Concentration=

[TiOH] =

⇒[TiOH] = 4.52 × 10^-3 M

Now, the ionisation of TiOH

⇒TiOH↔ Ti⁺ + OH^-

TiOH is completely ionised

∴ [TiOH] = [OH^-] = 4.52 × 10^-3 M

To calculate [H⁺], we apply the formula

K_w = [H⁺] [OH^-]

As the value of K_w is taken as 10^-14

∴ [H⁺] =

⇒[H⁺] = 2.21 × 10^-12 M

To calculate pH, we apply the formula:

pH = -log[H⁺]

⇒pH = - log (2.21 × 10^-12)

⇒pH = -log 2.21 – (-12) log 10

⇒pH = 12 – log2.21

⇒pH = 12 – 0.3424

⇒pH = 11.65

Thus, the pH of the given solution is 11.65

b) For 0.3g of Ca(OH)₂ dissolved in water to give 500mL of solution,

Given:

Mass of Ca(OH)₂ (m) = 0.3g

Volume(V) = 500 ml = L = 0.5L

Molar mass of Ca(OH)₂ (M) = 40 × 2(16 + 1) = 74 g/mol

To calculate concentration of Ca(OH)₂, we apply the formula,

Concentration=

⇒[Ca(OH)₂] =

⇒[Ca(OH)₂] = 8.1 × 10^-3 M

Now, the ionisation of Ca(OH)₂

⇒Ca(OH)₂↔ Ca²⁺ + 2OH^-

Ca(OH)₂ is completely ionised

∴ [Ca(OH)₂] = 2[OH^-]

⇒2[OH^-]= 2 × 8.1 × 10^-3 M

⇒16 × 10^-3 M

To calculate pOH, we apply the formula:

pOH = -log[OH^-]

⇒pOH = -log (16 × 10^-3)

⇒pOH = -log 16 – (-3) log 10

⇒pOH = 3 -1.201

⇒pOH = 1.79

As we know that pH + pOH = 14

∴ pH = 14 – 1.79

⇒pH = 12.21

Thus, the pH of the given solution is 12.21

c)for 0.3g of NaOH dissolved in water to give 500mL of solution,

Given:

Mass of NaOH (m) = 0.3g

Volume(V) = 200 ml = L = 0.2L

Molar mass of NaOH (M) = 40 g/mol

To calculate concentration of NaOH, we apply the formula,

Concentration=

⇒[NaOH] =

⇒[NaOH] = 3.75 × 10^-2 M

Now, the ionisation of NaOH

⇒NaOH↔ Na + OH^-

NaOH is completely ionised

∴ [NaOH] = [OH^-]

⇒[OH^-]= 3.75 × 10^-2 M

To calculate pOH, we apply the formula:

pOH = -log[OH^-]

⇒pOH = -log (3.75 × 10^-2 ) M

⇒pOH = -log 3.75 – (-2) log 10

⇒pOH = 2-log 3.75

⇒pOH = 1.42

As we know that pH + pOH = 14

∴ pH = 14 – 1.42

⇒pH = 12.57

Thus, the pH of the given solution is 12.57

d)For 1mL of 13.6 M HCl diluted with water to give 1 L of solution:

Given:

M₁ = 13.6M

V₁ = 1ml

V₂ = 1L = 1000ml

∴ 13.6 × 1 mL = M₂ × 1000 mL

⇒M₂ = 1.36 × 10^-2

Ionization of HCl

HCl ⇌ H⁺ + Cl-

As it is completely ionized

∴ [H⁺] = [HCl]

⇒1.36 × 10^-2

To calculate pH, we apply the formula:

pH = -log[H⁺]

⇒pH = - log (1.36× 10^-2)

⇒pH = -log 1.36 – (-2) log 10

⇒pH = 2 – log 1.36

⇒pH = 2 – 0.1335

⇒pH = 1.8665

Thus, the pH of the solution is 1.87