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SSSSS [86.1K]
3 years ago
8

DUPLICATE. The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to

form its conjugate, and the other half still remains unreacted. If 0.580 moles of a monoprotic weak acid (Ka = 4.5 _ 10-5) is titrated with NaOH, what is the pH of the solution at the half-equivalence point?
Chemistry
1 answer:
Law Incorporation [45]3 years ago
5 0
At the half equivalence point [HA] = [A-] and pH = pKa 

<span>if Ka is 5.2e-5 then pKa = pH = 4.28</span>
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A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s
Lilit [14]

Answer:

pH = 1.32

Explanation:

                 H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount  to buffer solutions.  In the first  deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g  =  0.074 mol H₂M

54 mL x  1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

           

3 0
3 years ago
Calculate the force applied to a car with mass of 1200 kg that has an acceleration of 2 m/s/s.
bearhunter [10]

Answer:

2,400

Explanation:

F = m × a

F = 1200 × 2

F = 2,400

6 0
3 years ago
A mixture of H2 and water vapor is present in a closed vessel at 20°C. The total pressure of the system is 755.0 mmHg.
Masja [62]

THE ANSWER IS: <u>737.5</u>

I JUST TOOK THE QUIZ!!!!

7 0
3 years ago
Read 2 more answers
2. Suppose that 21.37 mL of NaOH is needed to titrate 10.00 mL of 0.1450 M H2SO4 solution.
AysviL [449]

Answer:

0.1357 M

Explanation:

(a) The balanced reaction is shown below as:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

(b) Moles of H_2SO_4 can be calculated as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For H_2SO_4 :

Molarity = 0.1450 M

Volume = 10.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 10×10⁻³ L

Thus, moles of H_2SO_4 :

Moles=0.1450 \times {10\times 10^{-3}}\ moles

Moles of H_2SO_4  = 0.00145 moles

From the reaction,

1 mole of H_2SO_4 react with 2 moles of NaOH

0.00145 mole of H_2SO_4 react with 2*0.00145 mole of NaOH

Moles of NaOH = 0.0029 moles

Volume = 21.37 mL = 21.37×10⁻³ L

Molarity = Moles / Volume = 0.0029 /  21.37×10⁻³  M = 0.1357 M

7 0
3 years ago
How many protons electrons and neutrons does the isotope nitrogen 15 have
Ivenika [448]
Answer:

The atomic number of<span>N<span>157</span></span> 
The number of protons is 7
The number of electrons is 7
The number of neutrons is 8

Explanation:

The atomic number of Nitrogen is 7 because Nitrogen has 7 protons. 
The seven protons attract 7 electrons in the ground state.

If the atom had fewer or more than 7 protons the atom would not be Nitrogen.

The mass of the atom is the sum of protons and neutron. so

p + n = mass ( protons (p) and neutrons(n) both have an atomic mass of one

7 + n = 15 subtract 7 from both sides

<span>7−7+n=15−7</span>

n = 8

4 0
3 years ago
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