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
spin [16.1K]
2 years ago
7

A solution is prepared by mixing 250 mL of 1.00 M CH3COOH with 500 mL of 1.00 M NaCH3COO. What is the pH of this solution? (Ka f

or CH3COOH = 1.8 × 10−5 )
Chemistry
2 answers:
natta225 [31]2 years ago
7 0

Answer:

4.745

Explanation:

We are given from the question that 250 mL of 1.00 M CH3COOH and 500 mL of 1.00 M NaCH3COO makes a solution.

The equation of Reaction is given below;

CH3COOH + NaCH3COO ----> NaCH3COOH + CH3COONa.

==> Step 1 : Calculating the number of moles of each reactants and the find the limiting reagent.

One mole of CH3COOH reacts with one mole of NaCH3COO to give one NaCH3COOH and one mole CH3COONa.

Number of moles CH3COOH = Concentration of CH3COOH × volume of CH3COOH.

Number of moles CH3COOH= 1.00 M × 250 × 10^-3 litres.

Number of moles CH3COOH= 0.25 moles.

Number of moles of NaCH3COO = Concentration × volume.

Number of moles of NaCH3COO= 1.00 M × 500 × 10^-3 litres.

Number of moles of NaCH3COO= 0.5 moles.

Acetic acid is the limiting reagent. Therefore the numbers of moles of NaCH3COO that remains=( 0.5 - 0.25 )moles. = 0.25 moles.

=== > Step 2: use the formula in equation (1) below to calculate pH value.

First, the total volume of the solution =( 250 + 500)mL = 750 mL.

Thus, the Concentration of NaCH3COO= remaining moles/ total volume= 0.25 moles / 750 × 10^3 Litres.

Concentration of CH3COONa=0.25/ 0.75 = .33 M.

Concentration of CH3COOH = 0.25/ 0.75 = .33 M.

pH= pKa + log ([conjugate base]/ [weak acid] ) . --------------------(1).

pKa= - log ka.

pKa = - log 1.8 × 10^-5= 4.745.

Hence, pH= 4.745 + log ([.33)/ [.33]).

pH = 4.745 + 0.

pH = 4.745

Svetllana [295]2 years ago
5 0

Answer:

A solution is prepared by mixing 250 mL of 1.00 M

CH3COOH with 500 mL of 1.00 M NaCH3COO.

What is the pH of this solution?

(Ka for CH3COOH = 1.8 × 10−5 )

Explanation:

This is a case of a neutralization reaction that takes place between acetic acid,     CH 3 COOH ,   a weak acid, and sodium hydroxide,   NaOH , a strong base.

The resulting solution pH, depends if the neutralization is complete or not.  If not, that is, if the acid is not completely neutralized, a buffer solution containing acetic acid will be gotten, and its conjugate base, the acetate anion.

It's important to note that at complete neutralization, the pH of the solution will not equal  7 . Even if the weak acid is neutralized completely, the solution will be left with its conjugate base, this is the reason why the expectations of its pH is to be over  7 .

So, the balanced chemical equation for this reaction is the ionic equation:

CH 3 COOH (aq]  +  OH − (aq]  →  CH 3 COO − (aq]  +  H 2 O (l]

Notice that:  

1  mole of acetic acid will react with:  1  mole of sodium hydroxide, shown here as hydroxide anions,  OH − , to produce   1   mole of acetate anions:

CH 3 COO −

To determine how many moles of each you're adding , the molarities and volumes of the two solutions are used:

     c  =  n /  V    ⇒     n   =   c  ⋅  V

n  acetic   =   0.20 M   ⋅   25.00   ⋅   10  − 3 L   =   0.0050 moles CH3 COOH

and

n  hydroxide   =   0.10 M   ⋅   40.00   ⋅   10 − 3 L   =   0.0040 moles OH −

There are fewer moles of hydroxide anions, so the added base will be completely consumed by the reaction.

As a result, the number of moles of acetic acid that remain in solution is:

    n  acetic remaining   =   0.0050  −   0.0040   =    0.0010 moles

The reaction will also produce  0.0040   moles of acetate anions.

This is, then a buffer and the Henderson-Hasselbalch equation is applied to find its pH :

pH  =  p K a  +  log  ( [ conjugate base ]  / [ weak acid ] )

Use the total volume of the solution to find the new concentrations of the acid and of its conjugate base .

V total  =  V acetic  +  V hydroxide

V total  =  25.00 mL  +  40.00 mL  =  65.00 mL

Thus the concentrations will be :

[ CH 3 COOH ]  =  0.0010 moles  / 65.00  ⋅  10 − 3 L  =  0.015385 M

and

[ CH 3 COO − ]  =  0.0040 moles  / 65  ⋅  10 − 3 L  =  0.061538 M

The    p K a     of acetic acid is equal to    4.75

Thus the pH of the solution will be:

pH   =   4.75  +  log ( 0.061538 M  /    0.015385 M )

pH   =   5.35

You might be interested in
In the Lab you report a measured volume 128.7 of water. Using significant figures as a measure, what range of answers does your
Cerrena [4.2K]

Your measurement implies that the range of answers is 128.6 mL to 128.8 mL.

If you do not state explicitly the range of uncertainty (e.g., ± 0.5 mL), the <em>implied range of uncertainty</em> is ±1 in the last significant digit.

Thus, a reading of 128.7 mL implies that the volume is 128.7 mL ± 0.1 mL.

7 0
3 years ago
What is the molar mass of potassium bromide, KBR?
Nezavi [6.7K]

Answer:

id.............k

Explanation:

3 0
3 years ago
What is the formula for tin(IV) sulfide?<br> A. Sn4S<br> B. SnS2<br> C. Sns<br> D. SnS4
defon

Answer:

SnS_{2}

Explanation:

The formula for tin(IV) sulfide is SnS_{2}

6 0
2 years ago
The vapor pressure of pure water at 25 °c is 23.8 torr. What is the vapor pressure (torr) of water above a solution prepared by
maxonik [38]

The vapour pressure of the solution is 23.4 torr.

Use <em>Raoult’s Law</em> to calculate the vapour pressure:  

<em>p</em>₁ = χ₁<em>p</em>₁°  

where  

χ₁ = the mole fraction of the solvent  

<em>p</em>₁ and <em>p</em>₁° are the vapour pressures of the solution and of the pure solvent  

The formula for vapour pressure lowering Δ<em>p</em> is  

Δ<em>p</em> = <em>p</em>₁° - <em>p</em>₁  

Δ<em>p</em> = <em>p</em>₁° - χ₁<em>p</em>₁° = p₁°(1 – χ₁) = χ₂<em>p</em>₁°  

where χ₂ is the mole fraction of the solute.  

<em>Step 1</em>. Calculate the <em>mole fraction of glucose </em>

<em>n</em>₂ = 18.0 g glu × (1 moL glu/180.0 g glu) = 0.1000 mol glu  

<em>n</em>₁ = 95.0 g H_2O × (1 mol H_2O/18.02 g H_2O) = 5.272 mol H_2O  

χ₂ = <em>n</em>₂/(<em>n</em>₁ + n₂) = 0.1000/(0.1000 + 5.272) = 0.1000/5.372 = 0.018 62  

<em>Step 2</em>. Calculate the <em>vapour pressure lowering</em>  

Δ<em>p</em> = χ₂<em>p</em>₁° = 0.018 62 × 23.8 torr = 0.4430 torr  

<em>Step 3</em>. Calculate the <em>vapour pressure</em>  

<em>p₁</em> = <em>p</em>₁° - Δ<em>p</em> = 23.8 torr – 0.4430 torr = 23.4 torr

3 0
3 years ago
Rank the elements according to highest ionization energy:<br><br> Be, C, O, Ne, B, Li, F, N
aleksklad [387]
Given:
Be - Beryllium   -   9,3227
C   - Carbon      - 11,2603
O   - Oxygen      - 13,6181
Ne - Neon          - 21,5645
B   - Boron          -   8,298
Li  - Lithium        -   5,3917
F   - Fluorine      - 17,4228
N   - Nitrogen    - 14,5341

Arranged from highest ionization energy to lowest ionization energy.

Ne ; F ; N ; O ; C ; Be ; B ; Li
5 0
3 years ago
Read 2 more answers
Other questions:
  • About half of the electricity in the United States is generated using _____. natural gas biomass coal hydropower
    9·2 answers
  • Bromine has an atomic number of 35. How many protons are in an atom of bromine? 17 18 35 52
    5·1 answer
  • I have no clue pls help
    15·1 answer
  • Which of these statements best explains how fossil records provide evidence to the scientific theory of evolution?
    7·1 answer
  • Which state of matter expands when heated and is easy to compress?
    15·1 answer
  • What is 4.50+3.4+12.09 ?​
    14·2 answers
  • 2C(s) + 2H2(g) + 52.4kJ ⇋ C2H4(g)
    12·1 answer
  • The big bang theory states that​
    11·1 answer
  • 4.Which scientist described the electrons as traveling in certain orbits or energy levels?
    6·1 answer
  • A codon that stops the synthesis of a protein molecule
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!