Question

Determine the pH of the solution when 30.00 mL of 0.2947 M NH3 is titrated with 10.00 mL of 0.4798 M HCl.

Answer 1

no of moles of NH3 = 0.2947 mol/L x 0.03 L = 0.00884 moles

no o fmole of HCl = 0.4798 mol/L x 0.01 L = 0.0048

write the balanced equation

NH3 + HCl ----> NH4Cl

one mole of ammonia will react with one mole of HCl will produce on emole of NH4Cl

here no o fmoles of ammonia is more so all the HCl will consume

0.0048 moles of HCl will react with 0.00478 moles of NH3 will produce 0.0048 moles of NH4Cl

total volume = 30 +10 = 40 = 0.04L

now concentration of NH4Cl = no of moles of NH4Cl / volume of NH4Cl

= 0.0048 / 0.04 L

= 0.12 M

no of moles of NH3 remaining = 0.00884-0.0048 = 0.00404 molesNH3 remains

concentration NH3 remains = 0.00404 moles / 0.04 L = 0.101 M

now we have concentration of salt and concentration of NH3

pKb of NH3 = 9.25

use the handerson equation to find pOH

pOH = pKb + log(salt/base)

pOH = 9.25 + log(0.12/0.101)

pOH = 9.25 + 0.075

pOH = 9.325

pH = 14-pOH

pH = 14-9.325

pH = 4.675