We are given
0.2 M HCHO2 which is formic acid, a weak acid
and
0.15 M NaCHO2 which is a salt which can be formed by reacting HCHO2 and NaOH
The mixture of the two results to a basic buffer solution
To get the pH of a base buffer, we use the formula
pH = 14 - pOH = 14 - (pKa - log [salt]/[base])
We need the pKa of HCO2
From, literature, pKa = 1.77 x 10^-4
Substituting into the equation
pH = 14 - (1.77 x 10^-4 - log 0.15/0.2)
pH = 13.87
So, the pH of the buffer solution is 13.87
A pH of greater than 7 indicates that the solution is basic and a pH close to 14 indicates high alkalinity. This is due to the buffering effect of the salt on the base.
Answer:
ΔH₁₂ = -867.2 Kj
Explanation:
Find enthalpy for 3H₂ + O₃ => 3H₂O given ...
2H₂ + O₂ => 2H₂O ΔH₁ = -483.6 Kj
3O₂ => 2O₃ ΔH₂ = + 284.6 Kj
_____________________________
3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O (multiply by 3 to cancel O₂)
6H₂ + 3O₂ => 6H₂O ΔH₁ = 3(-483.6 Kj) = -1450.6Kj
2O₃ => 3O₂ ΔH₂ = -284.6Kj (reverse rxn to cancel O₂)
_______________________________
6H₂ + 2O₃ => 6H₂O ΔH₁₂ = -1735.2 Kj (Net Reaction - not reduced)
________________________________
divide by 2 => target equation (Net Reaction - reduced)
3H₂ + O₃ => 3H₂O ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj
Answer:
82.97 K
Explanation:
Applying,
PV/T = P'V'/T'................ Equation 1
Where P = initial pressure, T = Initial temperature, V = Initial Volume, P' = Final pressure, V' = Final Volume, T' = Final Temperature.
Make T' the subject of formula in equation 1
T' = P'V'T/PV................ Equation 2
From the question,
Given: P = 877 mmHg = (877×0.001316) atm = 1.154 atm, T = 222.2 K, V = 7.9 L, P' = 0.327 atm, V' = 10.41 L
Substitute these values into equation 2
T' = (0.327×10.41×222.2)/(1.154×7.9)
T' = 82.97 K
Hence the new temperature is 82.97 K
The picture isn’t included
