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mihalych1998 [28]
2 years ago
10

1.00 mL of 12.0 M HCl is added to 1.00 L of a buffer that contains 0.110 M HNO2 and 0.170 M NaNO2. How many moles of HNO2 and Na

NO2 remain in solution after addition of the HCl
Chemistry
1 answer:
eimsori [14]2 years ago
8 0

Answer:

Moles of NaNO2 = 0.158

Moles of HNO2 final = 0.098

Explanation:

Given

Moles of HCl = 12

Moles of HNO2 = 0.11

Moles of NaNO2 = 0.170

HCl +NaNO2 --> HNO2  + NaCl

1 mole of HCl react with one mole of NaNO2 to produce 1 mole of NaCl and 1 mole of HNO2

Moles of NaNO2 = 0.17 - 0.012 = 0.158

Moles of HNO2 final = 0.11 - 0.012 = 0.098

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<u>Answer:</u> The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.

<u>Explanation:</u>

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.

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A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

The given redox reaction follows:

MnO_4^-(aq)+NO_2^-(aq)\rightarrow MnO_2(s)+NO_3^-(aq)

To balance the given redox reaction in basic medium, there are few steps to be followed:

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Reduction half-reaction: MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-

  • Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions

Oxidation half-reaction: NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-         ( × 3)

Reduction half-reaction: MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-             ( × 2)

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Reduction half-reaction: 2MnO_4^-+4H_2O+3e^-\rightarrow 2MnO_2+8OH^-

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Overall redox reaction: 3NO_2^-+2MnO_4^-+H_2O\rightarrow 3NO_3^-+2MnO_2+2OH^-

As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution

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