Answer to Question #194614 in Chemistry for Arvin Shopon

Question

1.

a) What is the pH of a solution if it's oxonium ion concentration is 0.0025 mol/l ?

b) The pH of the solution is 10.4. Is the solution acidic or alkaline? What is the hydroxide ion concentration of the solution?

c) Determine the pH of the 0.035 molar potassium hydroxide solution.

Answer 1

Solution:

a)

oxonium ion = hydronium ion = H₃O⁺

[H₃O⁺] = 0.0025 mol/L

We can convert between [H₃O⁺] and pH using the following equation:

pH = - log[H₃O⁺]

pH = - log(0.0025) = 2.60

pH = 2.60

Answer (a): pH = 2.60

b)

pH = 10.4

The solution is alkaline (since pH > 7).

hydroxide ion = OH^-

For any aqueous solution at 25^∘C:

pH + pOH = 14

Thus, pOH = 14 - pH = 14 - 10.4 = 3.6

We can convert between pOH and [OH^-] using the following equation:

pOH = - log[OH^-]

[OH^-] = 10^-pH

[OH^-] = 10^-3.6 = 0.00025 mol/L

[OH^-] = 0.00025 mol/L

Answer (b): The solution is alkaline. [OH^-] = 0.00025 mol/L.

c)

potassium hydroxide = KOH

KOH is a strong base which dissociates as follows:

KOH → K⁺ + OH^-

[OH^-] = C(KOH) = 0.035 M (according to the equation above)

We can convert between [OH^-] and pOH using the following equation:

pOH = - log[OH^-]

pOH = - log(0.035 ) = 1.46

pOH = 1.46

For any aqueous solution at 25^∘C:

pH + pOH = 14

Thus, pH = 14 - pOH = 14 - 1.46 = 12.54

pH = 12.54

Answer (c): pH = 12.54