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assuming complete dissociation calculate the ph o...

Question:
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

Answer:

(i) 0.003MHCl:

H₂O + HCl ↔ H3O⁺ + Cl^-

Since HCl is completely ionized,

[H₃O⁺] = [ HCl]

⇒ [H₃O⁺] = = 0.003

Now

pH = -log [H₃O⁺] = -log (0.003)

= 2.52

Hence, the pH of the solution is 2.52.

(b) 0.005 M NaOH

NaOH_(aq) ↔ Na⁺_(aq) + HO^-_(aq)

[NaOH] = [ HO^-]

⇒ [ HO^-] = 0.05

pOH = -log[ HO^-] = -log (0.05)

= 2.30

∴ pH = 14 - 2.30 = 11.70

Hence, the pH of the solution is 11.70.

HBr + H₂O ↔ H₃O⁺ + Br^-

[HBr] = [H₃O⁺]

⇒ [H₃O⁺] = 0.002

∴ pH = -log [H₃O⁺] = -log (0.002)

= 2.69

Hence, the pH of the solution is 2.69.

(d) 0.002 M KOH

KOH_(aq) ↔ K⁺_(aq) + OH^-_(aq)

[OH^-] = [KOH]

⇒ [OH^-] = 0.002

Now pOH = -log[OH^-] = -log (0.002)

= 2.69

∴ pH = 14-2.69 = 11.31

Hence, the pH of the solution is 11.31.

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Question:
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

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Question: 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

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Question:
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