Question

A sample of 0.320 M hydroxylamine (HONH2) is reacted with 0.210 M HBr solution. How many mL of the HBr would be required to exactly neutralize 25.0 mL of the HONH2 solution? answer: 38.10mL. What would be the pH of the solution from part E?

Answer 1

hydroxylamine HONH2 is a weak base and it reacts with acid like HBr to give salt HONH3Br. The reaction is -

HONH2 + HBr -> HONH3Br

So, we need 1 mole of HBr for each mole of HONH2 to react to produce 1 mole of acidic salt HONH3Br (for complete neutralization of HONH2)

Lets say we need x ml of HBr to do this so,

equating millimoles of HONH2 in 25ml of 0.32 M solution to that of millimoles of HBr in x ml of 0.21 M solution

milimoles = molarity*volume in ml

0.32M*25ml = 0.21M*xml

So, x = 8/0.21 ml = 38.10 ml

millimoles of HONH3Br produced = millimoles of HONH2 reacted =  0.32M*25ml

= 8 millimoles

Now, the final volume of the solution is 25ml (initial) + 38.10(HBr added) = 63.1 ml

So, the molarity of salt HONH3Br = millimoles/total volume in ml

= 8millimoles/63.1ml = 0.127 M

HONH3Br will undergo salt hydrolysis to give

HONH3Br -> HONH3+ Br-

HONH3+ <-> HONH2 + H+

I 0.127 0 0

C -0.127x +0.127x +0.127x

E 0.127(1-x) 0.127x 0.127x

Kh = [H+][HONH2]/[HONH3+] -- Eqn1

Kh = 0.127x*0.127x/(0.127(1-x))

Kh = 0.127x²/(1-x) ---- Eqn2

Also,

HONH2 + H2O <-> HONH3+ + OH-

Kb = [OH-][HONH3+]/[HONH2] --- Eqn3

Multiplying Eqn1 and Eqn3 we get

Kh*Kb = Kw (Kw = [H+][OH-])

so, Kh = Kw/Kb = 10^-14/10^-7.97 (Kb = 10^-pKb pKb value taken from wikipedia)

Kh = 10^-6.03

From Eqn2 we get

Kh = 0.127x²/(1-x) = 10^-6.03

Assuming x<<1 and thus neglecting x with respect to 1

x-1 is roughly equal to 1 so,

0.127x² = 10^-6.03

x² = 10^-6.03/0.127

x = sqrt(10^-6.03/0.127)

x = 0.00271

So, H+ = 0.127*x = 0.127*0.00271

So, H+ = 0.000344

so pH = -log(H+) = -log(0.000344)

pH = 3.463