Acid/Base Equilibrium - Strong Acids/Bases and pH 1. Calculate the hydroxide ion

Question

Acid/Base Equilibrium - Strong Acids/Bases and pH

1. Calculate the hydroxide ion concentration of a 2.51×10-2 M solution of the strong acid CF3CO2H. Round your answer to 3 significant digits. (Constant = Kw=10-14)

2. Calculate the molarity (M) of a solution of the strong acid CF3CO2H if the pH is 3.79. (Constant = Kw=10-14)

3. Calculate the molarity (M) of a solution of the strong base Ba(OH)2 if the hydroxide ion concentration is 0.458 M. (Constant = Kw=10-14)

4. Calculate the pH of a 1.83×10-2 M solution of the strong acid HI. Round your answer to 3 significant digits. (Constant = Kw=10-14)

1. Calculate the hydroxide ion concentration of a 2.51×10-2 M solution of the strong acid CF3CO2H. Round your answer to 3 significant digits. (Constant = Kw=10-14)

2. Calculate the molarity (M) of a solution of the strong acid CF3CO2H if the pH is 3.79. (Constant = Kw=10-14)

3. Calculate the molarity (M) of a solution of the strong base Ba(OH)2 if the hydroxide ion concentration is 0.458 M. (Constant = Kw=10-14)

4. Calculate the pH of a 1.83×10-2 M solution of the strong acid HI. Round your answer to 3 significant digits. (Constant = Kw=10-14)

Explanation / Answer

1) CF3COOH(aq) ----------> CF3COO-(aq) + H+(aq)

Since the acid is strong , therefore it will completely dissociate into ions in aqueous solution.Thus,

[H+] = [CF3COOH]initially = 2.51*10-2 M

2) pH = -log[H+]

Thus, [H+] = 10-pH = 1.622*10-4

CF3COOH(aq) ----------> CF3COO-(aq) + H+(aq)

Since the acid is strong , therefore it will completely dissociate into ions in aqueous solution.Thus,

[H+] = [CF3COOH]initially = 1.622*10-4 M

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