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
seraphim [82]
3 years ago
5

A 25.0 ml sample of an unknown monoprotic acid was titrated with 0.12 m naoh. the student added 31.6 ml of naoh and went past th

e equivalence point. which procedure could be performed next to more accurately determine the concentration of the unknown acid?
Chemistry
1 answer:
Inessa05 [86]3 years ago
7 0
The procedure, which can be used to determine more accurately the concentration of the unknown acid is TO BACK-TITRATE WITH ADDITIONAL HYDROCHLORIC ACID TO NEUTRALIZE THE ADDITIONAL SODIUM HYDROXIDE THAT WAS ADDED.
Monoprotic acids are acids that can donate only one proton per each molecule and they have only one equivalence point. Examples of monoprotic acids are HCI, HNO3 and CH3COOH.
The back titration method is typically used when one needs to determine the concentration of an analyte provided there is a known molar concentration of excess reactants. 
From the information given in the question above, we are told that excess NaOH was added. To correct this mistake, the right thing to do is to use additional HCl to carry out back titration, taking note of the quantity of acid that will be needed to neutralize the excess NaOH.
You might be interested in
A histidine is involved in an interaction with a glutamic acid that stabilizes the charged form of the histidine, such that the
Greeley [361]

Answer:

pKa of the histidine = 9.67

Explanation:

The relation between standard Gibbs energy and equilibrium constant is shown below as:

\Delta{G^0} =-RT \ln \frac{[His]}{[His+]}

R is Gas constant having value = 0.008314 kJ / K mol  

Given temperature, T = 293 K

Given, \Delta{G^0}=15\ kJ/mol

So,  Applying in the equation as:-

15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}

Thus,

15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}

\frac{[His]}{[His+]}=e^{\frac{15}{-0.008314\times 293}

\frac{[His]}{[His+]}=0.00211

Also, considering:-

pH=pKa+log\frac{[His]}{[His+]}

Given that:- pH = 7.0

So, 7.0=pKa+log0.00211

<u>pKa of the histidine = 9.67</u>

8 0
3 years ago
Due tonight! Please help!
olchik [2.2K]

Answer:

42.98 g/L

Explanation:

6 0
3 years ago
Ethanol (c2h5oh) melts at -114°c. the enthalpy of fusion is 5.02 kj/mol. the specific heats of solid and liquid ethanol are 0.9
Nimfa-mama [501]

Answer: 4.18925 kJ heat is needed to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.

Explanation:

Temperature of Solid C_2H_5OH=-135^oC=138 K(0^oC=273K)

Melting temperature of Solid C_2H_5OH=114^oC=159 K

Temperature of liquid C_2H_5OH=-50^oC=223K

Specific heats of solid  ethanol = 0.97 J/gK

Specific heats of liquid ethanol = 2.3 J/gK

Heat required to melt the the 25 g solid C_2H_5OH at 159 K

\Delta T_1 = 159 K - 138 K = 21 K

Q_1=mc\Delta T= 25\times 0.97J/gK\times 21 K=509.25 J

Heat required to melt and raise the temperature of C_2H_5OH upto 223 K

\Delta T_2 = 223 K - 159 K  = 64 K

Q_2=mc\Delta T= 25\times 2.3J/gK\times 64 K=3680 J

Total heat to convert solid ethanol to liquid ethanol at given temperature :

Q_1+Q_2=509.25 J+3680 J=4189.25 J=4.18925 kJ (1kJ=1000J)

Hence, 4.18925 kJ of heat will be required to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.

6 0
3 years ago
A compound with chemical formula na2cx3 has formula mass 106 amu .. what is the atomic mass of element x
maria [59]

The element "X" is "O" (oxygen).

<h3>Calculation:</h3>

Given,

Chemical formula = Na₂CX₃

Formula mass = 106 amu

Molar mass of Na = 23 amu

Molar mass of C = 12 amu

To find,

Element X =?

We will equate the equation as follows,

2(23) + 12 + 3(y) = 106

46 + 12 + 3y =106

58 + 3y = 106

3y = 106 - 58

3y = 48

y = 48/3

y = 16

We know that Oxygen has molecular mass of 16. Therefore the element "X" is "O".

Learn more about molar mass here:

brainly.com/question/22997914

#SPJ4

5 0
1 year ago
To what volume should 25ml of 15m nitric acid be diluted to prepare a 3m solution
Maurinko [17]

We can use the dilution formula to find the volume of the diluted solution to be prepared

c1v1 = c2v2

Where c1 is concentration and v1 is volume of the concentrated solution

And c2 is concentration and v2 is volume of the diluted solution to be prepared

Substituting the values in the equation

15 M x 25 mL = 3 M x v2

v2 = 125 mL

The 25 mL concentrated solution should be diluted with distilled water upto 125 mL to make a 3 M solution

6 0
3 years ago
Other questions:
  • Which of the following is true for all chemical reactions
    8·1 answer
  • How many mol of lithium are there in 1.204 x 1024 lithium atoms
    11·1 answer
  • Explain evaporation. have a chapter test tommarow !
    12·2 answers
  • An 18.5 g sample of tin (molar mass = 118.7) combines with
    5·1 answer
  • You have an unknownthat contains only one cation (it is oneof the twelve from this experiment).Your solution is colorless and od
    7·1 answer
  • Please help im rlly stuck
    8·2 answers
  • PLEASE HELP me create a fictional video game using at least 2 of the ecological roles. (roles to choose from: decomposer, detrit
    12·1 answer
  • Answer correctly for a brainleist! :)
    5·1 answer
  • What is bond stability??​
    12·2 answers
  • What's the pH of a solution with a concentration of hydronium ions of 5.29x10^-10?
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!