Titration of NH3 with HCl

Question

Titrate 10 cm3 of 0.1 M NH3 solution with 0.1 M HCl solution (factor= 0.998). After how many cm3 of HCl have been added, the pH of the oldst will be 8.8?

Answer 1

"oldst"?

NH₃ + HCl ==> NH₄Cl

moles NH₃ initially present = 10 cm3 x 1 1 L/1000 cm3 x 0.1 mol/L = 0.001 mol NH₃

When adding HCl to NH₃, you form a buffer solution since you have a weak base (NH₃) and the conjugate acid (NH₄⁺). Using a form of the Henderson Hasselbalch equation, we can solve for the [conj.acid]/[base] ratio as follows:

pOH = pKb + log[conj.acid]/[base] and pOH = 14 - pH = 14 - 8.8 = 5.2

pOH = 5.2 = 4.75 + log x

log x = 0.45

x = [NH₄⁺] / [NH₃] = 2.8

NH₃ ===> NH₄⁺

0.001...........0........initial

-x...............+x.........change

0.001-x.......x........equilibrium

x / 0.001-x = 2.8

x = 7.4x10^-4 mole = moles NH₃ reacted = moles HCl to be added

volume HCl needed: (x L)(0.1 mol/L) = 7.4x10^-4 mol and x = 7.4x10^-3 L = 7.4 ml

Use the ICE table to check this answer:

NH3 + HCl ===> NH4Cl

0.001.......0.00074...............0........initial

-0.00074...-0.00074....+0.00074 ... change

0.00026....0..............0.00074........equilibrium

NH₄⁺ / NH₃ = 0.00074 / 0.00026 = 2.8 = desired ratio calculated above