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suter [353]
3 years ago
9

You titrate 25.00 ml of 0.1894 M acetic acid with 0.2006 M NaOH. How many ml of NaOH (to four significant figures) will be requi

red at the equivalence point?
Chemistry
1 answer:
Anvisha [2.4K]3 years ago
8 0

Answer:

23.60 mL NaOH

Explanation:

The reaction is CH3COOH + OH- --> CH3COO+ + H2O

Since the reaction is one-to-one, we can use M1V1 = M2V2.

M1 = 0.1894 M CH3COOH

V1 = 25.00 mL CH3COOH

M2 = 0.2006 M NaOH

V2 = ?

Solve for V2 --> V2 = M1V1/M2

V2 = (0.1894 M)(25.00 mL) / (0.2006 M) = 23.60 mL NaOH

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Student carries out a titration to determine the concentration of a solution of
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<h3>Further explanation  </h3>

Titration is a procedure for determining the concentration of a solution (analyte) by reacting with another solution whose known concentration (usually a standard solution) is called the titrant. Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range  

Titrations can be acid-base titration, depositional titration, and redox titration. An acid-base titration is the principle of neutralization of acids and bases  

Reaction

HNO₃ + NaOH → NaNO₃ + H₂O​

Concentration a standard solution  of sodium hydroxide :  0.0998 mol/dm³ , and the volume = 25 cm³

moles NaOH=

\tt mol=M\times V\\\\mol=0.0998\times 25\\\\mol=2.495~mlmoles

<em>From the equation, mol ratio HNO₃ : NaOH = 1 : 1, so mol HNO₃ = mol NaOH=</em><em>2.495 mlmoles</em>

<em></em>

The volume of HNO₃ = 21.8 cm³, so the concentration :

\tt M=\dfrac{n}{V}\\\\M=\dfrac{2.495}{21.8}\\\\M=0.114

7 0
2 years ago
A concentration cell consisting of two hydrogen electrodes (PH2 = 1 atm), where the cathode is a standard hydrogen electrode and
Lunna [17]

The pH of the unknown solution is 3.07.

<u>Explanation:</u>

<u>1.Find the cell potential as a function of pH</u>

From the Nernst Equation:

Ecell=E∘cell−RT /zF × lnQ

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R denotes the Universal Gas Constant

T denotes the temperature

z denotes the moles of electrons transferred per mole of hydrogen

F denotes the Faraday constant

Q denotes the reaction quotient

Substitute the values,

E∘cell=0   lnQ=2.303logQ

E0cell=−2.30/RT /zF × log Q

Solving the equation,

<u>2. Find the Q  value</u>

Q=[H+]2prod pH₂, product/ [H+]2reactpH₂, reactant

Q=[H+]^2×1/1×1=[H+]2

Taking the log

logQ= log[H+]^2=2log[H+]=-2pH

From the formula,

Ecell=−2.303RT /zF× logQ

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K × 2pH /2×96 485  C⋅mol

( inverse)

E cell= 0.0592 V × pH

<u>3. Finding the pH value</u>

E cell= 0.0592 V × pH

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pH=3.07

The pH of the unknown solution is 3.07.

7 0
3 years ago
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