Question

1. Short Answer: Consider the titration of 50.0 mL of 20 M HNO. (K. - 4.0x10' with 1.0 MNOH. Fill in pH values in this table for each amount of strong base added to weak acid (25 points) 1.0 M NaOH O mL Moles of NaOH added Calculated pll (2 sig fig) 25 ml 50 ml. 100 ml 150 ml Before filling in above table construct an ICE chart for each titration step: construct your own on following blank page, cample provided here, and below each ICE chart SHOW ALL WORK for your calculations of pH at each step. Reactant ! Reactant 2 Product ! Product 2 (if needed) b. Use your data from part a to sketch a rough graph of titration with pH on y axis and V of NaOH on x axis. Use the graph to identify the midpoint and the equivalence point of the titration (15 points) 25 50 75 100 125 150 Based on this data, what would be the most effective pH buffering range (you may justify your answer with shape of graph and/or Henderson-Hasselbalch equation) (10 points)

Answer 1

II.(a):

0 mL NaOH:

Moles of NaOH added = 0 mol

--HNO₂(aq) <-----> NO₂^-(aq) + H⁺​​​​​​​(aq): Ka = 4.0*10^-4

I: 2.00 M --------------- 0 M ------- 0 M

C: - X ------------------- +X -------- +X

E:(2.00 - X) ------------ X ---------- X

Ka = 4.0*10^-4  = [NO₂^-(aq)]*[H⁺​​​​​​​(aq)] / [HNO₂​​​​​​​(aq)]

=> 4.0*10^-4  = X*X / (2.00 - X)

=> X² = 8.0*10^-4 - 4.0*10^-4 *X

Solving the above quadratic equation gives:

=> X = 0.0281 M

=> X =[H⁺​​​​​​​(aq)] = 0.0281 M

=> calculated pH = - log[H⁺​​​​​​​(aq)] = - log(0.0281) = 1.55 (Answer)

# (b):

25 mL NaOH:

Moles of NaOH added = C*V = 1.0 mol/L * 0.025 L = 0.025 mol

Initial moles of HNO₂​​​​​​​(aq) = C*V = 2.0 mol/L * 0.050 L = 0.100 mol

-- HNO₂​​​​​​​(aq) + NaOH(aq) ----> NaNO₂(aq) + H2O(l)

I: 0.100 mol, 0.025 mol, --------0  mol

C: - 0.025 mol, -0.025 mol ----- + 0.025 mol

E: (0.100-0.025), (0.100-0.025), 0.025

= 0.075 mol, ----- 0 mol ------------ 0.025 mol

HNO₂​​​​​​​(aq) and NaNO₂(aq) acts as buffer solution. Hence

Applying henderson equation

pH = pKa + log([NaNO₂(aq)] / [HNO₂​​​​​​​(aq)])

Since volume for both NaNO₂(aq) and HNO₂​​​​​​​(aq) is same, we can replace concentration by mole.

pH = pKa + log(moles of NaNO₂(aq) / moles of HNO₂​​​​​​​(aq))

=> pH = - log(4.0*10^-4) + log(0.025 / 0.075)

=> Calculated pH = 2.92 (Answer)

# (c):

50 mL NaOH:

Moles of NaOH added = C*V = 1.0 mol/L * 0.050 L = 0.050 mol

Initial moles of HNO₂​​​​​​​(aq) = C*V = 2.0 mol/L * 0.050 L = 0.100 mol

Here moles of NaOH added ha exactly half of (0.1/2 = 0.050) the initial moles of HNO₂​​​​​​​(aq).

Hence this half-equivalence point.

Hence pH = pKa = - log(4.0*10^-4)

=> Calculated pH = 3.40

# (d):

100 mL NaOH:

Moles of NaOH added = C*V = 1.0 mol/L * 0.100 L = 0.100 mol

Initial moles of HNO₂​​​​​​​(aq) = C*V = 2.0 mol/L * 0.050 L = 0.100 mol

Since equal amount of both are added, this is equivalence point.

Total volume = 0.100 L + 0.050 L = 0.150 L

All HNO₂​​​​​​​(aq) converts to NaNO₂​​​​​​​(aq)

=> [NaNO₂​​​​​​​(aq)] = 0.100 mol / 0150L = 0.6667 M

NaNO₂​​​​​​​(aq) undergo hydrolysis to form HNO₂​​​​​​​(aq) and OH^-(aq).

--NO₂^-​​​​​​​(aq) + H2O(l) <-----> HNO₂​​​​​​​(aq) + OH^-​​​​​​​(aq): Kb = Kw/Ka = 10^-14 / 4.0*10^-4 = 2.5*10^-11

I: 0.667 M ------------------------ 0 M --------- 0 M

C: - X ---------------------------- +X ---------- +X

E:(0.667 - X) ---------------------- X ------------ X

=> Kb = 2.5*10^-11 = X*X / (0.667 - X)

Since X << 0.667, we can neglect X in denominator.

=> 2.5*10^-11= X²

=> X = [OH^-​​​​​​​(aq)] = 5.0*10^-6 M

=> pOH = - log(5.0*10^-6) = 5.30

=> calculated pH = 14 - 5.30 = 8.70