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Volgvan
1 year ago
13

When a 27.7 mL sample of a 0.400 M aqueous hypochlorous acid solution is titrated with a 0.335 M aqueous potassium hydroxide sol

ution, what is the pH at the midpoint in the titration
Chemistry
1 answer:
Harlamova29_29 [7]1 year ago
4 0

With a 27.7 mL sample of a 0.400 M aqueous hypochlorous acid solution  titrated with a 0.335 M aqueous potassium hydroxide solution, the pH value is mathematically given as

pH=12.90

<h3>What is the pH at the midpoint in the titration?</h3>

Moles of HClO=0.329*21.5

Moles of HClO=7.0735

and

Moles of KOH=0.456*23.3

Moles of KOH=10.6248

Hence

The net mole of KOH=10.6248-7.0735

The  net mole of KOH=3.5513

Volume of solution=21.5+23.3

Volume of solution=44.8ml

Generally, the equation for the  Concentration of KOH is mathematically given as

CKOH=mole/volume

Therefore

CKOH=3.5513/44.8ml

CKOH=0.07927M

In conclusion,

pOH=-log{OH-}

pOH=-log{0.07927}

pOH=1.10

pH=14-110

pH=12.90

Read more about pH value

brainly.com/question/15845370

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Answer:

the experimental variable

Explanation:

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A chemist prepares a solution of barium chloride by measuring out of barium chloride into a volumetric flask and filling the fla
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The given question is incomplete. The complete question is:

A chemist prepares a solution of barium chloride by measuring out 110 g of barium chloride into a 440 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mole per liter of the chemist's barium chloride solution. Round your answer to 3 significant digits.

Answer: Concentration of the chemist's barium chloride solution is 1.20 mol/L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n\times }{V_s}

where,

n = moles of solute

V_s = volume of solution in L

moles of BaCl_2(solute) = \frac{\text {given mass}}{\text {Molar Mass}}=\frac{110g}{208g/mol}=0.529mol

Now put all the given values in the formula of molality, we get

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Answer:

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Ag⁺(aq) + 1 e⁻ → Ag(s)

We can establish the following relations.

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The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

19.0s \times \frac{0.770c}{s} \times \frac{1mole^{-} }{96,468C} \times \frac{1molAg}{1mole^{-}} \times \frac{107.87g}{1molAg} = 0.0164 g

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<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The overall chemical reaction follows:

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The intermediate balanced chemical reaction are:

(1) Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\rightarrow CaCO_3(s)    \Delta H_1=-812.8kJ  

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The expression for enthalpy of the reaction follows:

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Hence, when the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed.

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Answer:

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