Question

A 150 mL of 0.15 M NH3 is slowly added to 25 mL of 0.2 M HCl. What is the pH of the resulting solution? Is this final solution buffered?

Answer 1

no of moles of NH3 = Molarity of NH3 x volume of NH3 in Liters = 0.15M x 0.15L = 0.0225 mol

no ofmoles of HCl = Molarity of HCl x volume of HCl = 0.2M x 0.025L = 0.005 mol

now write the balanced equation

NH3 + HCl -----> NH4Cl

from this balanced equation one mole of ammonia will react with one mole of HCl will give one mole of NH4Cl

here moles of HCl is less so

0.005 moles of HCl will react with 0.005 moles of NH3 and will produce 0.005 moles of NH4Cl

no of moles of NH3 remains = 0.0225 - 0.005 =0.0175 moles

total volume = 150 + 25 = 175 mL = 0.175L

concentration of NH4Cl = 0.005 / 0.175 = 0.0286M

concentration of NH3 remaining = 0.0175 / 0.175 = 0.1M

now use the handerson equation

pOH = pKb + log[salt/base]

pKb of NH3 = 4.74 from wiki

pOH = 4.74 - log[0.0286/0.1]

pOH = 4.74 - 0.54

pOH = 4.2

pH = 14-pOH = 14-4.2

= 9.8