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

i believe it is c.) 1575 j.

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.