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belka [17]
3 years ago
12

Consider a buffer solution that is 0.50 M in NH3 and 0.20 M in NH4Cl. For ammonia, pKb=4.75. Calculate the pH of 1.0 L of the so

lution upon addition of 30.0 mL of 1.0 M HCl to the original buffer solution.
Express your answer to two decimal places.
***will mark Brainliest for right answer ***
Chemistry
1 answer:
vladimir2022 [97]3 years ago
5 0
You have 0.50 mol of NH3 and 0.20 mol of NH4+ to start (NH4Cl dissolves completely), given the molarity and 1.0 L solution.

30.0 mL of 1.0 M HCl is 0.0300 mol of HCl. This will react with the NH3 to produced 0.030 mol of NH4+.

You now have 0.47 mol NH3 and 0.23 mol NH4+. Now use the Henderson-Hasselbach equation to calculate your pH. The equation says to use concentration of acid and base, but you can just use the moles of them because it doesn’t make a difference.

pH = pKa + log(base/acid)

pKa = 14 - pKb = 14 - 4.75 = 9.25

pH = 9.25 + log(0.47/0.23) = 9.56
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The reduction or addition of electrons will be occurring in cations and the oxidation or removal of electrons will be occurring in anions.

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3 years ago
222 g of MTBE (CO(CH3)4) are added to gasoline, resulting in a total volume of 2 L of reformulated gas (RFG). Assume that the de
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Answer:

Explanation:

From the information given:

(a)

Concentration \ in \ mg/L = \dfrac{Mass \ of \ MTBE \ in \ mg}{Total \ volume (in \ L)}

Concentration \ in \ mg/L = \dfrac{222 \times 10^3 \ mg}{22}

Concentration \ in \ mg/L = 111 \times 10^3 \ mg/L

(b)

number \ of \ mole s= \dfrac{mass}{molar \ mass } \\ \\ number \ of \ mole s=\dfrac{222 \ g}{88.15 \ g/mol} \\ \\ \mathbf{= 2.518 mol}

(c)

w/w \ percentage = \dfrac{mass \ of \ MTBE }{mass \ of \ solution (RFG)}\times 100\%

where; \\ \\ mass \  of \  (RFG) = 2L \times 0.70 g/mL \\ \\ mass \  of \  (RFG) = 2000 ml \times 0.70 g/mL \\ \\ mass \ of \ (RFG) = 1400 g

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w/w \ percentage = \dfrac{222 \ g}{1400 \ g}\times 100\% = \mathbf{15.8\%}

(d)

Volume of MTBE =\dfrac{mass \ of \ MTBE}{density \ of \ MTBE}

Volume \ of \ MTBE = 300 \ mL\\

∴

v/v\% = \dfrac{volume \ of \ MTBE}{volume \ of \ RFG} \\ \\ v/v\% =\dfrac{300 \ mL}{2000 \ mL}\times 100\% \\ \\ \mathbf{v/v\% = 15.00\%}

(e)

From \the  \ given \  information; \\ \\ 2.5184 \ moles\ of  \ MTBE contain  \ 2.5184  \ mole of oxygen

∴

mass of oxygen MTBE = 2.5284 mol \times 16\ g/mol \\ \\ mass of oxygen MTBE = 40.3 9 \ g\\ \\ mass\ of \ RFG = 1400 g

∴

\% w/w = \dfrac{mass \ of \ oxygen}{mass \ of RFG }=\dfrac{40.22 \ g}{1400 \ g} \times 100\%

\% w/w == 2.88\%

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3 years ago
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<h3>What is a balanced Reaction ?</h3>

A reaction in which the number of atoms in reactants is equal to the number of atoms in the products is called a balanced reaction.

The reaction given in the question is

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Adding the equation will give balance equation

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for 1 mole of S₂O₃²⁻  0.5 mole of S₂O₈²⁻  will be used.

To know more about Balanced Reaction

brainly.com/question/14280002

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