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Strike441 [17]
3 years ago
15

Four acids are analyzed and their Ka values are determined. Which of the following values represents the strongest acid?

Chemistry
2 answers:
amm18123 years ago
6 0

Answer:

D) Ka=1.2•10^-2

Explanation:

#platolivesmatter

labwork [276]3 years ago
4 0
Hi, thank you for posting your question here at Brainly. 

The acid dissociation constant, Ka, is an equilibrium constant that measure the strength of an acid. It is a ratio of the concentration of the products (salt and water) and the reactants (acid). The higher the Ka, the more tendency it is to favor the product side, which means more tendency to donate H+ ions. This is exactly the definition of a strong acid (high H+ ionized).

Thus, the answer is letter D.
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Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr
VikaD [51]

Answer : The value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

Explanation :

The balanced cell reaction will be,

Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)

The half-cell reactions are:

Oxidation reaction (anode) : Pb(s)\rightarrow Pb^{2+}(aq)+2e^-

Reduction reaction (cathode) : 2Ag^+(aq)+2e^-\rightarrow 2Ag(g)

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

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

\Delta G^o=-2\times 96500\times 0.93

\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol

Now we have to calculate the value of 'K'.

\Delta G^o=-RT\ln K

where,

\Delta G_^o =  standard Gibbs free energy  = -180 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K

K=3.6\times 10^{31}

Therefore, the value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

5 0
3 years ago
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