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3 years ago

Which one of these salts will form a basic solution on dissolving in water?

Chemistry

1 answer:

ipn [44]3 years ago

6 0

Answer:

KCN forms a basic solution (option B)

Explanation:

Step 1: Data given

Acids have form AH, with conjugate base A-.

Bases have the form BOH, with conjugate acid B+.

Strong acids have weak conjugate bases

Strong bases have weak conjugate acids

Weak acids have strong conjugate bases

Weak base have strong conjugate acids

a) NaCl.

Na+ is a conjugate acid of NaOH, what is a strong base.

Cl- is the conjugate base of HCl, what is a strong acid.

They form a neutral solution

b) KCN.

K+ is a conjugate acid of KOH, what is a strong base.

CN- is the conjugate base of HCN, what is a weak acid.

K+ is a weak conjugate acid, CN- is a stronger conjugate base

They form a basic solution

c) NaNO3.

Na+ is a conjugate acid of NaOH, what is a strong base.

NO3- is the conjugate base of HNO3 , what is a strong acid.

They will form a neutral solution

d) NH4NO3.

NH4+ (ammonium) is the conjugate acid of NH3, what is a moderately strong base.

NO3- is the conjugate base of HNO3, what is a strong acid.

Ammonium is a stronger conjugate acid, nitrate is a weak conjugate base.

The solution is acid.

e) FeCl3 is an acid salt because Fe(OH)3 is weak base and HCl is a strong acid. pH is less than 7.

Fe^3+ is the conjugate acid of Fe(OH)3, what is weak base.

Cl- is the conjugate base of HCl, what is a strong acid.

Fe^3+ is a 'stronger' conjugate acid, Cl- is a weaker conjugate base.

They solution is acid

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3 years ago

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Answer : The ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln Q\\\\\Delta G^o=-RT\times \ln (\frac{[A]_{inside}}{[A]_{outside}})$

where,

$\Delta G^o$ = standard Gibbs free energy = -14.1 kJ/mol

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$[A]_{outside}$ = concentration outside the cell

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

$-14.1\times 10^3J/mol =-(8.314J/K.mol)\times (298K)\times \ln (\frac{[A]_{inside}}{[A]_{outside}})$

$\frac{[A]_{inside}}{[A]_{outside}}=296.2$

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Explanation:

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