Question

What is the pH of an aqueous solution made by combining 35.01 mL of a 0.3791 M ammonium chloride with 40.07 mL of a 0.3542 M solution of ammonia to which 3.112 mL of a 0.0896 M solution of HCI was added?

Answer 1

pH = 9.25

Step-by-step explanation:

Using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where pH is pH of the buffer

pKa is pKa of the weak acid: 9.24

[A⁻] could be taken as moles of the conjugate base: Moles NH₃

[HA] moles of the weak acid: NH₄⁺

Initial moles of NH₃:

40.07mL = 0.04007L * (0.3542mol / L) = 0.01419mol

Initial moles NH₄⁺:

0.03501L * (0.3791mol / L) = 0.01327 mol

The HCl added reacts with NH₃ producing NH₄⁺:

HCl + NH₃ → NH₄⁺ + Cl⁻

Moles of HCl:

3.112x10⁻³L * (0.0896mol / L) = 0.00028 moles

That means moles of NH₃ and NH₄⁺ after the addition of HCl are:

NH₄⁺: 0.01327moles + 0.00028moles = 0.01355 moles

NH₃: 0.01419mol - 0.00028 moles = 0.01391 moles

pH of the buffer is:

pH = pKa + log [A⁻] / [HA]

pH = 9.24 + log [0.01391 moles] / [0.01355 moles]

pH = 9.25