Question

In: Chemistry

12. Consider the titration of 40.0 mL of 0.500 M NH3 with 1.00 M HCl a)...

12. Consider the titration of 40.0 mL of 0.500 M NH3 with 1.00 M HCl a) What is the initial pH of the NH3(aq)? b) What is the pH halfway to the equivalence point? c) What is the volume of HCl needed to reach the equivalence point? d) What is the pH at the equivalence point? e) Sketch the titration curve. Label the point(s) where there is a A) a weak base B) weak acid C) Buffer D) Strong acid in excess

Expert Solution

a)

This is a base in water so, let the base be CH3NH2 = "B" and CH3NH3+ = HB+ the protonated base "HB+"

there are free OH- ions so, expect a basic pH

B + H2O <-> HB+ + OH-

The equilibrium Kb:

Kb = [HB+][OH-]/[B]

initially:

[HB+] = 0

[OH-] = 0

[B] = M

the change

[HB+] = x

[OH-] = x

[B] = - x

in equilibrium

[HB+] = 0 + x

[OH-] = 0 + x

[B] = M - x

Now substitute in Kb

Kb = [HB+][OH-]/[B]

Kb = x*x/(M-x)

x^2 + Kbx - M*Kb = 0

x^2 + (1.8*10^-5)x - (0.5)(1.8*10^-5= 0

solve for x

x = 0.00299

substitute:

[HB+] = 0 + x = 0.00299M

[OH-] = 0 + x = 0.00299M

pH = 14 + pOH = 14 + log(0.00299) = 11.48

b)

pH half way is given by

pH = pKa + log(NH3/NH4+)

pKa = 14 ´Kb = 14-4.25 = 9.25

pH = 9.25+ log(1)

pH = 9.25

c)

V o fHcl required

mmol of base = MV = 40*0.5 = 20

mmol of acid = 20

V acid =mol/M = 20/1 = 20 mL required

d)

pH in equivalence

[NH4+] = 20/(40+20) = 0.33

the next equilibrium is formed, the conjugate acid and water

BH+(aq) + H2O(l) <-> B(aq) + H3O+(aq)

The equilibrium is best described by Ka, the acid constant

Ka by definition since it is an base:

Ka = [H3O][B]/[BH+]

Ka can be calculated as follows:

Ka = Kw/Kb = (10^-14)/(1.8*10^-555) = 5.55*10^-10

get ICE table:

Initially

[H3O+] = 0

[B] = 0

[BH+] = M

the Change

[H3O+] = 0 + x

[B] = 0 + x

[BH+] = -x

in Equilibrium

[H3O+] = 0 + x

[B] = 0 + x

[BH+] = M - x

substitute in Kb expression

Kb = [HA ][OH-]/[A-]

5.55*10^-10 = x*x/(0.333-x)

solve for x

x^2 + Ka*x - M*Ka= 0

solve for x with quadratic equation

x = H3O+ =1.359*10^-5

[H3O+] =1.359*10^-5

pH = -log([H3O+]) = -log(1.359*10^-5) = 4.87

pH = 4.87

queen_honey_blossom answered 2 years ago

Consider the titration of 40.00 mL of 0.200 M NH3 with 0.500 M HCl. Calculate pH...

Consider the titration of 40.00 mL of 0.200 M NH3 with 0.500 M HCl. Calculate pH a) initially b) after addition of 5.00 mL of 0.500 M HCl c) at the half-neutralization point (halfway to equilvalence point) d) after the addition of 10.00 mL of 0.500 M HCl e) at the equivalence point

Consider the titration of 40.0 mL of 0.243-M of KX with 0.097-M HCl. The pKa of...

Consider the titration of 40.0 mL of 0.243-M of KX with 0.097-M HCl. The pKa of HX = 7.09. Give all pH values to 0.01 pH units. a) What is the pH of the original solution before addition of any acid? pH = b) How many mL of acid are required to reach the equivalence point? VA = mL c) What is the pH at the equivalence point? pH = d) What is the pH of the solution after the addition...

Consider a titration involving 40.0 mL of 0.100 M ammonia , NH3 (in an Erlenmeyer flask)with...

Consider a titration involving 40.0 mL of 0.100 M ammonia , NH3 (in an Erlenmeyer flask)with 0.100 M HCl (in a burette) to answer the next two questions. NH3 + HCl

Consider the titration of 30.0 mL of 0.030 M NH3 with 0.025 M HCl. Calculate the...

Consider the titration of 30.0 mL of 0.030 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added. (a) 35.0 mL (b) 36.0 mL (c) 37.0 mL

2. For the titration of 20 mL of 0.1026 M NH3 with 0.09747 M HCl, calculate...

2. For the titration of 20 mL of 0.1026 M NH3 with 0.09747 M HCl, calculate the pH before the addition of titrant, at 10 mL prior to the equivalence point, at the equivalence point, and at 10 mL after the equivalence point. Ka (NH4+) = 5.7 x 10-10

A student conducts a titration of 50 mL of HCl with 1.00 M NaOH. The pH...

A student conducts a titration of 50 mL of HCl with 1.00 M NaOH. The pH at the equivalence point is Basic Neutral Acid

Consider the titration of 0.100 M HCl, a strong acid, with 0.100 M NH3, a weak...

Consider the titration of 0.100 M HCl, a strong acid, with 0.100 M NH3, a weak base: HCL(aq)+NH3(aq)---NH4+(aq)+H2O(l) Calculate the pH at the following points in the titration when 25.0 mL of 0.0100 M NH3 is titrated with 0.100 M HCL. A. 0.0 mL of HCl added B. 10.0 mL of HCl added C. At equivalence point D. 35.0 mL of HCL added

Consider the titration of 40.0 mL of 0.200 M HClO4 by 0.100 M KOH. Calculate the...

Consider the titration of 40.0 mL of 0.200 M HClO4 by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. a. 0.0 mL d. 80.0 mL b. 10.0 mL e. 100.0 mL c. 40.0 mL answers are A) 0.699 B) 0.854 C) 1.301 D) 7.00 E) 12.15 I would just like to know how to do the work please and tbank you

calculate the pH for the titration 20 mL of 0.500 M HF with 0.500 M KOH...

calculate the pH for the titration 20 mL of 0.500 M HF with 0.500 M KOH at the following volumes ka, HF is 6.8E-4 using the Rice table a.0 mL b.10 mL c.19 mL d. 20 mL e. 21 mL f. 30 mL

Consider the titration of 40.0 mL of 0.523 M ascorbic acid (H2C6H6O6) with 0.885 M potassium...

Consider the titration of 40.0 mL of 0.523 M ascorbic acid (H2C6H6O6) with 0.885 M potassium hydroxide (KOH): pKa1 = 4.10 pKa2 = 11.80 What is the solution pH before the titration begins? _________________ What is the pH at the first half-equivalence point? What is the pH at the first equivalence point? _________________ _________________ What is the pH at the second half-equivalence point? _________________ What is the pH at the second equivalence point? _________________ What is the pH when 70.0...