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?

pH = 9.25

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]

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.