Calculate the pH of 20.00 mL of 0.150 M of a weak acid after 15.00 mL of 0.200 M NaOH is added to the solution. The Ka of the weak acid is 1.80 X 10-5. Write the answer in 3 significant figures. g

pH = 8.84

Explanation:

The reaction of HA + NaOH is:

HA + NaOH → H₂O + Na⁺ + A⁻

20.00mL of 0.150M HA are:

0.020L × (0.150mol / L) = 3.00x10⁻³ moles of HA

Moles of NaOH are:

0.015L × (0.150mol / L) = 3.00x10⁻³ moles of NaOH

That means all HA reacts producing 3.00x10⁻³ moles A⁻. In 35mL of total volume: 3.00x10⁻³ moles A⁻ / 0.035L = 0.0857M.

The equilibrium of A⁻ in water is:

A⁻ + H₂O ⇄ HA + OH⁻

Where Kb is (Kw / Ka =  1.00x10⁻¹⁴ / 1.80x10⁻⁵ = 5.56x10⁻¹⁰) defined as:

Kb = [HA] [OH⁻] / [A⁻]

Equilibrium concentrations are:

[A⁻] = 0.0857M - X

[HA] = X

[OH⁻] = X

Replacing in Kb formula:

5.56x10⁻¹⁰ = [X] [X] / [0.0857M - X]

4.77x10⁻¹¹ - 5.56x10⁻¹⁰X = X²

0 = X² + 5.56x10⁻¹⁰X - 4.77x10⁻¹¹

Solving for x:

X = -6.907x10⁻⁶ M  → False answer → There is no negative concentrations

X = 6.906x10⁻⁶ M → Right answer

As X = [OH⁻] and pOH = -log [OH⁻], pOH =-log 6.906x10⁻⁶ M = 5.16

As 14 = pH + pOH

pH = 14 - 5.16

pH = 8.84

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.