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3 years ago

What is the pH of 1.00 L of a buffer that is 0.110 M nitrous acid (HNO2) and 0.200 M NaNO2? (pKa of HNO2 = 3.40)

Chemistry

1 answer:

Andrei [34K]3 years ago

4 0

Answer:

pH = 3.65

Explanation:

given data

pKa of HNO2 = 3.40

nitrous acid (HNO2) = 0.110 M

NaNO2 = 0.200 M

to find out

What is the pH

solution

we get here ph for acidic buffer that is express as

pH = pKa + log(salt÷acid) ........................1

put here value and we get

pH = 3.40 + log(0.200÷0.110)

pH = 3.65

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For b: The standard Gibbs free energy of the reaction is 746.91 kJ

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Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)

For a:

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

Oxidation half reaction: $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

Reduction half reaction: $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

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$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

For b:

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

Oxidation half reaction: $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

Reduction half reaction: $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

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