Assuming complete dissociation, calculate the pH of the following solutions:

A. 0.003 M HCl B. 0.005 M NaOH

C. 0.002 M HBr D. 0.002 M KOH

To calculate the pH, we apply the formula:

For acidic solution, pH = -log[H₃O⁺]

For basic solution, pOH = - log[OH^-]

Also, we use the given formula to calculate pH and pOH

pH + pOH = 14

A)0.003M HCl

Given:

[HCl] = 0.003 M

Ionization of HCl

HCl + H₂O ⇌ H₃O⁺ + Cl-

Since HCl is completely ionized(given)

[HCl] = [H₃O⁺]

⇒[H₃O⁺] = 0.003 M

By applying the formula,

pH = -log[H₃O⁺]

As [H₃O⁺] = 0.003 M

∴ pH = -log[0.003]

⇒pH = 2.52

Thus, the pH of the solution is 2.52

B) 0.005 M NaOH

Given:

[NaOH] = 0.005M

Ionization of NaOH,

NaOH ⇌ Na⁺ + OH-

Since NaOH is completely ionized (given)

[NaOH] = [OH^-]

⇒[OH^-] = 0.005 M

By applying the formula for basic solution:

pOH = -log[OH^-]

As [OH^-] = 0.005 M

∴ pOH = -log [0.005]

⇒pOH = 2.30

As pOH + pH =14

⇒pH = 14 - pOH

⇒pH = 14 – 2.30

⇒pH = 11.70

Thus, the pH of the solution is 11.70

c)0.002 M HBr

Given:

[HBr] = 0.002 M

Ionization of HBr

H₂O + HBr ⇌ H₃O⁺ + Br-

Since HBr is completely ionized(given)

[HBr] = [H₃O⁺]

⇒[H₃O⁺] = 0.002 M

By applying the formula (for acidic solution)

pH = -log[H₃O⁺]

As [H₃O⁺] = 0.002 M

∴ pH = -log[0.002]

⇒pH = 2.69

Thus, the pH of the solution is 2.69

d)0.002 M KOH

Given:

[KOH] = 0.002M

Ionization of KOH

KOH + ⇌ K⁺ + OH-

Since KOH is completely ionized (given)

[KOH] = [OH^-]

⇒[OH^-] = 0.002 M

By applying the formula for basic solution,

pOH = -log[OH^-]

As [OH^-] = 0.002 M

∴ pOH = -log[0.002]

⇒pOH = 2.69

As pOH + pH =14

⇒pH = 14 - pOH

⇒pH = 14 – 2.69

⇒pH = 11.31

Thus, the pH of the solution is 11.31