Answer : The pH of a 0.1 M phosphate buffer is, 6.86
Explanation : Given,
Concentration of acid = 0.1 M
Concentration of conjugate base (salt) = 0.1 M
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
Now put all the given values in this expression, we get:
Therefore, the pH of a 0.1 M phosphate buffer is, 6.86
Answer : The value of equilibrium constant for this reaction at 328.0 K is
Explanation :
As we know that,
where,
= standard Gibbs free energy = ?
= standard enthalpy = 151.2 kJ = 151200 J
= standard entropy = 169.4 J/K
T = temperature of reaction = 328.0 K
Now put all the given values in the above formula, we get:
The relation between the equilibrium constant and standard Gibbs free energy is:
where,
= standard Gibbs free energy = 95636.8 J
R = gas constant = 8.314 J/K.mol
T = temperature = 328.0 K
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:
Therefore, the value of equilibrium constant for this reaction at 328.0 K is
Answer:
2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.
Explanation:
Moles of NaOH =
Molarity of the nitric acid solution = 0.250 M
Volume of the nitric solution = 0.150 L
Moles of nitric acid = n
According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :
of NaOH
Moles of NaOH left unreacted in the solution =
= 0.375 mol - 0.0375 mol = 0.3375 mol
1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.
Then 0.3375 moles of NaOH will give :
of hydroxide ion
The molarity of hydroxide ion in solution ;
2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.