1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
patriot [66]
3 years ago
9

Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result

ing solution after each of the following volumes of NaOH has been added. (Assume that all solutions are at 25°C.)(a) 0.0 mLWebAssign will check your answer for the correct number of significant figures. (b) 50.0 mLWebAssign will check your answer for the correct number of significant figures. (c) 100.0 mLWebAssign will check your answer for the correct number of significant figures. (d) 150.0 mLWebAssign will check your answer for the correct number of significant figures. (e) 200.0 mLWebAssign will check your answer for the correct number of significant figures. (f) 250.0 mLWebAssign will check your answer for the correct number of significant figures.
Chemistry
1 answer:
Murljashka [212]3 years ago
4 0

Answer:

(a) 2.7

(b) 4.44

(c) 4.886

(d) 5.363

(e) 5.570

(f)  12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only  the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) =  4.886

(d) mol HA reacted = 150 x 10⁻³ L  x  x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) =  5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH  = √(kb x (A⁻)  where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA,  200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒   Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

You might be interested in
Question 5 (1 point)<br> Potassium Bromide is insoluble.<br> O True<br> O False
Delicious77 [7]
Is soluble in water but not soluble in acetonitrile.

So i think its false
4 0
3 years ago
Read 2 more answers
Can someone please help me
mrs_skeptik [129]
The answer will be D
8 0
3 years ago
Analysis of the water content of a lake found in the desert showed that it contained 17.5 percent chloride ion, and has a densit
tankabanditka [31]

Answer:

The molarity is 6.0 M.

Explanation:

<em>The </em><em>concentration</em><em> of a solution is the amount of solute present in a given amount  of solvent, or a given amount of solution.</em>

In this problem, we are asked to convert percent by mass to molarity, which are two different concentration units.

<em>The </em><em>percent by mass </em><em>is the ratio of the mass of a solute to the mass of the solution, multiplied by 100 percent.</em> On the other hand, <em>molarity (M)</em><em> is defined as the number of moles of solute per liter of solution.</em>

If we have 17.15 percent chloride ion, that means that we have 17.15 grams of chloride ion in 100 grams of solution. The density provided is the density of the solution, so we calculate how many mL correspond to 100 grams of solution:

1.23 g ----------- 1 mL

100 g ------------ <u>x= 81.3 mL</u>

Therefore, 17.5 grams of chloride ion are contained in 81.3 mL. Now we need to convert these grams into moles.

The atomic mass of chlorine is 35.5 g/mol:

35.5 g --------------- 1 mol

17.5 g --------------- <u>x= 0. 49 mol</u>

This 0.49 moles are contained in 81.13 mL, the definition of molarity says that this moles are contained in a liter (or what is the same, 1000 mL):

81.3 mL --------------- 0.49 mol

1000 mL ------------- x= 6.0 M

The molarity of the chloride ion will be 6.0 M.

5 0
3 years ago
What is the molarity of the solution resulting from the dissolution of 239 g glucose (C6H12O6) in 250
kifflom [539]

Answer:

Molarity =5.32 M

Explanation:

Given data:

Mass of glucose = 239 g

Volume = 250 mL (250 /1000 = 0.25 L)

Molarity = ?

Solution;

Formula:

Molarity = number of moles / volume in litter

Number of moles:

Number of moles = mass/ molar mass

Number of moles = 239 g / 180.2 g/mol

Number of moles = 1.33 mol

Molarity:

Molarity = number of moles / volume in litter

Molarity = 1.33 mol / 0.25 L

Molarity =5.32 M

6 0
3 years ago
What is the average velocity of atoms in 1.00 mol of argon (a monatomic gas) at 275 K? For m, use 0.0399 kg. (1 point)
valentinak56 [21]

What is the average velocity of atoms in 1.00 mol of argon (a monatomic gas) at 275 k for m, use 0.0399kg

Answer: The average velocity of the atoms 847.33 m/s.

Explanation:

Moles of the neon = 1.00

Temperature of the gas : 288 K

Mass of the gas = 0.01000

R = 8.31 J/mol K

The average velocity of the atoms 847.33 m/s.

6 0
3 years ago
Other questions:
  • In order from LEAST density to most in order
    10·1 answer
  • What elements fill the d orbitals on the periodic table?
    11·1 answer
  • why should the pka of a leaving groups conjugate acid be considered when performing a substitution reaction
    6·1 answer
  • Each of the following substrates can react with a nucleophile in a substitution reaction. Select the substrate that cannot under
    13·1 answer
  • Where does reduction occur in a battery
    5·1 answer
  • A 10-gram sample of zinc loses 560 J of heat and has a final temperature of 100
    15·1 answer
  • What is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen?
    14·1 answer
  • Is crumpling a piece of paper chemical changes
    7·2 answers
  • How many moles are found in a 100.0 g sample of CuF2?
    8·2 answers
  • A 3L sample of an ideal gas at 178 K has a pressure of 0.3 atm. Assuming that the volume is constant, what is the approximate pr
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!