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Snezhnost [94]
4 years ago
9

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:

(Assume that the volume does not change upon the addition of calcium hydroxide.)
a. Raise the pH slightly
b. Lower the pH slightly
c. Raise the pH by several units
d. Lower the pH by several units
e. Not change the pH
f. Exceed the buffer capacity
Chemistry
1 answer:
Inessa05 [86]4 years ago
4 0

Answer:

a. Raise the pH slightly

Explanation:

We know that

Pka of HNO2/KNO2 =3.39

Moles of HNO2 in the buffer=0.247 mol/L×1L=0.247 moles

Moles of NO2-=0.329mol/L×1L=0.329 moles

If 0.271 moles of Ca(OH)2 is added it will neutralise 0.136 moles of acid ,HNO2,remaining HNO2=0.247-0.136=0.111 moles

Moles of NO2- will increase as 0.0333 moles Ca(NO)2 will be formed =0.0333+0.036=0.0693 moles

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

=3.39+log (0.0693/0.0317)=3.39+0.34=3.73

pH=3.73

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notsponge [240]

<u>Answer:</u> The concentration of Sn^{2+} in the cell is 9.0\times 10^{-3}M

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u>  Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-   E^o_{Zn^{2+}/Zn}=-0.76V

<u>Reduction half reaction:</u>  Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)   E^o_{Sn^{2+}/Sn}=-0.136V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.136-(-0.76)=0.624V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}

where,

E_{cell} = electrode potential of the cell = 0.660 V

E^o_{cell} = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

[Zn^{2+}]=2.5\times 10^{-3}M

[Sn^{2+}] = ?

Putting values in above equation, we get:

0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})

[Sn^{2+}]=9.0\times 10^{-3}M

Hence, the concentration of Sn^{2+} ions is 9.0\times 10^{-3}M

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