Question

Calculate the pH of the solution that results when mixing 20.0ml of 0.1750M HCl with 25.0ml of a_distilled water b_0.132M NaOH c_0.132M NH3 d_0.232M NaOH

Answer 1

a) pH = -lg [H + ]
C M = n/V
n (HCl) = 0.1750 · 20.0 = 3.5 mmol
C M (HCl) = 3.5 / 45.0 = 0.08 M
pH = -lg [0.08] = 1.1
b) HCl + NaOH = NaCl + H 2 O
n (HCl) = n (NaOH)
n (HCl) = 0.1750 · 20.0 = 3.5 mmol
n (NaOH) = 0.132 · 25 = 3.3 mmol
Here HCl is in excess: n (HCl) = 3.5 – 3.3 = 0.2 mmol
C M (HCl) = 0.2 / 45.0 = 0.004 M
pH = -lg [0.004] = 2.4
c) HCl + NH 3 = NH 4 Cl
n (HCl) = n (NH 3 )
n (HCl) = 0.1750 · 20.0 = 3.5 mmol
n (NH 3 ) = 0.132 · 25 = 3.3 mmol
Here HCl is in excess: n (HCl) = 3.5 – 3.3 = 0.2 mmol
C M (HCl) = 0.2 / 45.0 = 0.004 M
pH = -lg [0.004] = 2.4
d) HCl + NaOH = NaCl + H 2 O
n (HCl) = n (NaOH)
n (HCl) = 0.1750 · 20.0 = 3.5 mmol
n (NaOH) = 0.232 · 25 = 5.8 mmol
Here NaOH is in excess: n (NaOH) = 5.8 – 3.5 = 2.3 mmol
C M (NaOH) = 2.3 / 45.0 = 0.05 M
pOH = -lg [0.05] = 1.3
pH = 14 – 1.3 = 12.7