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3 years ago

What is the concentration of nitrate ions in a 0.225 msr(no3)2 solution?

Chemistry

2 answers:

12345 [234]3 years ago

7 0

Answer : The concentration of nitrate ions is 0.450 m.

Solution : Given,

Molarity of $Sr(NO_3)_2$ = 0.225 m

The balanced ionic equation is,

$Sr(NO_3)_2\rightarrow Sr^{+2}+2NO^-_3$

From the reaction, we conclude that the 1 mole of $Sr(NO_3)_2$ produces 2 moles of $NO^-_3$ ions. That means the concentration of nitrate ions is twice the value of $Sr(NO_3)_2$ .

then, 0.225 m of $Sr(NO_3)_2$ gives 2 × 0.225 of $NO^-_3$ ions

Now the concentration of $NO^-_3$ ions is equal to 0.450 m.

BabaBlast [244]3 years ago

5 0

We are given that there are 2 moles of NO3 from the 0.225 M Sr(NO3)2 solution. Therefore we simply use stoichiometry to solve this problem.

Concentration NO3 = 0.225 M Sr(NO3)2 * (2 moles NO3 / 1 mole Sr(NO3)2)

Concentration NO3 = 0.45 M

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A 1 liter solution contains 0.247 M nitrous acid and 0.329 M sodium nitrite. Addition of 0.271 moles of calcium hydroxide will:

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Answer:

a. Raise the pH slightly

Explanation:

We know that

Pka of HNO2/KNO2 =3.39

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pH=pka+log [base]/[acid] {henderson -hasselbach equation}

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pH=3.73

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Sindrei [870]

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Calculation ,

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