A 25.0 mL solution of quinine was titrated with 1.00 M hydrochloric acid, HCl. It was found that the solution contained 0.125 moles of quinine. What was the pH of the solution after 50.00 mL of the HCl solution were added

pH = 9.08

Explanation:

Quinine, C₂₀H₂₄O₂N₂, Q, is a weak base that, in water, has as equilibrium:

Q + H₂O ⇄ QH⁺ + OH⁻

Where pKb is 5.10

Using H-H equation for weak bases:

pOH = pKb + log₁₀ [QH⁺] / [Q]

The reaction of quinine with HCl is:

Q + HCl → QH⁺ + Cl⁻

Initial moles of quinine are 0.125 moles and moles added of HCl are:

0.05000L × (1.00mol / L) = 0.05000moles.

That means after the addition of 50.00mL of the HCl solution, moles of Q and QH⁺ are:

Q = 0.125mol - 0.050mol = 0.075 moles

QH⁺ = 0.050 moles

Replacing in H-H equation:

pOH = 5.10 + log₁₀ [0.050] / [0.075]

pOH = 4.92

As pH = 14 - pOJ

it should be p1 v1 = p2 v2

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.