The pH of a solution is 9.02.
c(HCN) = 1.25 M; concentration of the cyanide acid
n(NaCN) = 1.37 mol; amount of the salt
V = 1.699 l; volume of the solution
c(NaCN) = 1.37 mol ÷ 1.699 l
c(NaCN) = 0.806 M; concentration of the salt
Ka = 6.2 × 10⁻¹⁰; acid constant
pKa = -logKa
pKa = - log (6.2 × 10⁻¹⁰)
pKa = 9.21
Henderson–Hasselbalch equation for the buffer solution:
pH = pKa + log(cs/ck)
pH = pKa + log(cs/ck)
pH = 9.21 + log (0.806M/1.25M)
pH = 9.21 - 0.19
pH = 9.02; potential of hydrogen
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Answer:
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Answer:
a H2CO3 b HCO3- and c H+ and HCO3-
Explanation:
As the pKa value of phenol is more than that of carbonic acid(H2CO3), the carbonic acid will have high Ka value than that of phenol.
The acid that contain high Ka value act as stong acid.From that point of view H2CO3 is a strong acid than phenol as the Ka value of carbonic acid is greater than that of phenol.
The conjugate base of H2CO3 is bicarbonate ion(HCO3-)
c The species that predorminates at equilibrium are H+ and HCO3-
Answer:
V = 85.619 L
Explanation:
To solve, we can use the ideal gas law equation, PV = nRT.
P = pressure (645 mmHg)
V = volume (?)
n = amount of substance (3.00 mol)
R = ideal gas constant (62.4 L mmHg/mole K)
T = temperature (295K)
Now we would plug in the appropriate numbers into the equation using the information given and solve for V.
(645)(V) = (3.00)(62.4)(295)
(V) = (3.00)(62.4)(295)/645
V = 85.619 L
