Answer to Question #207934 in General Chemistry for Earl

Question

Solve the following problem.

1. The concenration of H+ ions in a bottle of table wine was 3.2 x 10^-4 M right after the cork was removed. Only half of the wine was consumed. The other half, after it had been standing open to the air for a month, was found to have a H+ ion concentration equal to 1.0 x 10^-3 M. Calculate the pH of the wine on each of these occasions.

2. The pH of rainwater collected in ANHS on particular day was 4.82. Calculate the H+ ion concentration of the rainwater.

3. Find the pH if the [OH-] is equal to 1.0 x 10^-6 M.

Answer 1

1).

• Hⁱ_f=3.2 ×10^-4 M is the initial hydrogen ion concentration
• Hf+=1 × 10^-3M is the final hydrogen ion concentration

pH=−log[H⁺]

for the recently opened wine, we have:

pH_i=- log[3.2 × 10^-4 M]

pH_i = 3.49

As for the one month open wine,

pHf=−log[1 × 10^-3 M]

pH_f=3

2). pH = -log (H⁺)

4.82 = -log (H⁺)

H⁺ = 0.683

= 6.8 × 10^-1

3). PH = -log(H⁺) =

pOH = log (OH^-)

= -Log(1.0 × 10^-6)

= 6

pH + pOH = 14

pH = 14-6

= 8