Question

A 130.0 mL buffer solution is 0.105 molL−1 in NH3 and 0.125 molL−1 in NH4Br. What mass of HCl will this buffer neutralize before the pH falls below 9.00?If the same volume of the buffer were 0.270 molL−1 in NH3 and 0.395 molL−1 in NH4Br, what mass of HCl could be handled before the pH fell below 9.00?

Answer 1

We know that,

pKb for NH3 = 4.75

[NH3] = 0.105 M

[NH4Br] = 0.125 M

pH falls below 9 => pOH becomes greater than (14-9) = 5

pOH of the basic buffer = pKb + log ([NH4Br] / [NH3])

Let us suppose we added 'X' M of HCl

pOH after adding HCl = pKb + log ([NH4Br] + X / [NH3] - X)

Substituting limiting condition pOH = 5

5 = 4.75 + log (0.125 + X / 0.105 - X )

=> log (0.125 + X / 0.105 - X ) = 0.25

=> (0.125 + X / 0.105 - X ) = 1.778

=> X = 0.0222 M = [HCl]

Volume of solution = 130 mL = 0.13 L

=> Moles of HCl = 0.0222 x 0.13 = 0.0029 moles

Molar Mass of HCl = 36.5 g / mol

=> Mass of HCl = 0.0029 x 36.5 = 0.105 g HCl

Case 2)

5 = 4.75 + log (0.395 + X / 0.270 - X )

=> (0.395 + X / 0.270 - X ) = 1.778

=> X = 0.0306 M

=> Mass of HCl = 0.0306 x 36.5 = 1.12 g