A)when 0.0 mL of NaOH is added
HClO dissociates as:
HClO -----> H+ + ClO-
6*10^-2 0 0
6*10^-2-x x x
Ka = [H+][ClO-]/[HClO]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((2.9*10^-8)*6*10^-2) = 4.171*10^-5
since c is much greater than x, our assumption is correct
so, x = 4.171*10^-5 M
use:
pH = -log [H+]
= -log (4.171*10^-5)
= 4.3797
Answer: 4.38
B)when 12.5 mL of NaOH is added
Given:
M(HClO) = 0.06 M
V(HClO) = 50 mL
M(NaOH) = 0.12 M
V(NaOH) = 12.5 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.06 M * 50 mL = 3 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.12 M * 12.5 mL = 1.5 mmol
We have:
mol(HClO) = 3 mmol
mol(NaOH) = 1.5 mmol
1.5 mmol of both will react
excess HClO remaining = 1.5 mmol
Volume of Solution = 50 + 12.5 = 62.5 mL
[HClO] = 1.5 mmol/62.5 mL = 0.024M
[ClO-] = 1.5/62.5 = 0.024M
They form acidic buffer
acid is HClO
conjugate base is ClO-
Ka = 2.9*10^-8
pKa = - log (Ka)
= - log(2.9*10^-8)
= 7.538
use:
pH = pKa + log {[conjugate base]/[acid]}
= 7.538+ log {2.4*10^-2/2.4*10^-2}
= 7.538
Answer: 7.54
C)when 25.0 mL of NaOH is added
Given:
M(HClO) = 0.06 M
V(HClO) = 50 mL
M(NaOH) = 0.12 M
V(NaOH) = 25 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.06 M * 50 mL = 3 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.12 M * 25 mL = 3 mmol
We have:
mol(HClO) = 3 mmol
mol(NaOH) = 3 mmol
3 mmol of both will react to form ClO- and H2O
ClO- here is strong base
ClO- formed = 3 mmol
Volume of Solution = 50 + 25 = 75 mL
Kb of ClO- = Kw/Ka = 1*10^-14/2.9*10^-8 = 3.448*10^-7
concentration ofClO-,c = 3 mmol/75 mL = 0.04M
ClO- dissociates as
ClO- + H2O -----> HClO + OH-
0.04 0 0
0.04-x x x
Kb = [HClO][OH-]/[ClO-]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((3.448*10^-7)*4*10^-2) = 1.174*10^-4
since c is much greater than x, our assumption is correct
so, x = 1.174*10^-4 M
[OH-] = x = 1.174*10^-4 M
use:
pOH = -log [OH-]
= -log (1.174*10^-4)
= 3.9302
use:
PH = 14 - pOH
= 14 - 3.9302
= 10.0698
Answer: 10.07
D)when 40.0 mL of NaOH is added
Given:
M(HClO) = 0.06 M
V(HClO) = 50 mL
M(NaOH) = 0.12 M
V(NaOH) = 40 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.06 M * 50 mL = 3 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.12 M * 40 mL = 4.8 mmol
We have:
mol(HClO) = 3 mmol
mol(NaOH) = 4.8 mmol
3 mmol of both will react
excess NaOH remaining = 1.8 mmol
Volume of Solution = 50 + 40 = 90 mL
[OH-] = 1.8 mmol/90 mL = 0.02 M
use:
pOH = -log [OH-]
= -log (2*10^-2)
= 1.699
use:
PH = 14 - pOH
= 14 - 1.699
= 12.301
Answer: 12.30
Show ALL your calculations and the respective dissociation equations for each step. Assignments without calculations or...
Show ALL your calculations and the respective dissociation equations for each step. Assignments without calculations or assignments that are submitted late are not going to be graded. 1) Calculate the pH in the titration of 50.00 mL of 0.060 M hypochlorous acid (HCIO) with a 0.120 M sodium hydroxide, NaOH solution after the addition of the following volumes of base: Ka for hypochlorous acid = 2.9 x 10-8 A) 0 mL pH = pH = _ 7.5 B) 12.5 mL...
1) Calculate the pH in the titration of 50.00 mL of 0.060 M acetic acid (CH3COOH) with a 0.120 M sodium hydroxide, NaOH solution after the addition of the following volumes of base: Ka for acetic acid = 1.8 x 10-5 A) 0 mL pH = B) 10 ml pH =
please answer 23 and 24
*please show all work and calculations *
given temperature foart em to reach equilibrium e con 15) Pure Solic and pure liquide are excluded from equilibrium constant pre Cheric £69 kg 19) If a reaction is endothermic, the reaction temperature results in an ineren 20) A solution of ammonia is 2.0% ionized at 25.0 °C. What was the original concentration in M) of the ammonia solution? The Kb at 25.0 °C for ammonia is 1.8...
Calculations The following calculations are necessary to complete your Report Sheet. Consult the table on page 119 for appropriate K and K, values. Complete these calculations on your Report Sheet for complete credit. I. pH of Strong and Weak Acids 1. Calculate the pH of a 0.10 M hydrochloric acid solution, HCl(aq). 2. Calculate the pH of a solution prepared by diluting 5.00 mL of 0.10 M hydrochloric acid, HCl (aq), in enough water to make a 50.0 mL solution....
1. For each of the following, predict whether an aqueous solution of the salt would be acidic, basic or neutral. Explain. a. Nacis b. NaF c. NHIS d. KHCO3 e. (CH3)2NH2NO31 2. What is the pH of a solution that is 0.060 M in potassium propionate (CH3CH2COOK) and 0.085M in propionic acid (CH3CH2COOH)? The Ka for propionic acid is 1.3x10-5.5 3. Suppose you need to prepare a buffer for a pH of 10.6. An acid-base pair (HA/A') to use has...
please show all work for part III :) all the information
needed is in the pictures
III. Preparing HC,HO, Solutions and Determining pH only cal. One! theoretical pH 1.0 x 10-14 (181) concentration of HC,H,O, M measured pH ya ? 3.06 1.0 x 10-2 ka: 1. 8x10°3 ixio- 3.43 1.0 x 10-3 x = CH) -1,34% 103n 3.87 1.0 x 107 PM - 04 TH11 4.30 concentration of HC,H,O, M calculated K of HC H302 based on pH data literature...
**CHEMISTRY EXPERT ONLY**
please solve i will rate
For problems involving calculations you must show your work for credit. Unless otherwise stated, may assume T- 25.0 °C in all problems. you 1) Which of the following reactions goes essentially to completion? a) Reaction of a strong acid with a strong base b) Reaction of a strong acid with a weak base c) Reaction of a weak acid with a strong base d) All of the above e) None of the...
53) A buffer solution contains carbonic acid (H CO) and sodium bi carbonate (NaHCO), each at a concentration of 0.100 M. The relevant equilibrium is shown below. What is H:COs(ag)+ HOo (a)+HCOs (ag) K.-4.5x 10 54) At 40 C, the pH of water is 6.77, what is lon-Prodact Constant for Water, Kw.at 40°C 55) If a solution of acetic acid has a pH-2.74 and a pK474, wh in the solution? Show your calculations is the ratio of (CH Co/ICOH) 56)...
Please Show all work . Use attached Data and graph to answer
questions
Titration of Acetic Acid: 0. IM 1) Check to make sure that the volume on If it does not read 0.00 mL, adjust it to do so before you proceed any further. your sodium hydroxide buret reads 0.00 mL 2) Use the buret that is located near the acetic acid container to dispense 25.00 mL of acetic acid solution into a clean dry 100 mL beaker. Record...
52) (a) Use the Henderson-Hasselbalch equation to calculate the pt of NH CI and 0.15 M in NHb. (b) How would you prepare an NIH-CI-NHs buffer that has a pll of 9.00 (The Ka for NH, is 1.8 x 10) 53) A buffer solution contains carbonic acid (H:COs) and sodium bicarbonate (NaliCO,), each at a buffer solution that is 045 Mi concentration of 0.100 M. The relevant equilibrium is shown below, What is the pll of this buffer solutiont 54)...