<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>
Given:
Acetic Acid/Sodium Acetate buffer of pH = 5.0
Let HA = acetic acid
A- = sodium acetate
Total concentration [HA] + [A-] = 250 mM ------(1)
pKa(acetic acid) = 4.75
Based on Henderson-Hasselbalch equation
pH = pKa + log[A-]/[HA]
[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77
[A-] = 1.77[HA] -----(2)
From (1) and (2)
[HA] + 1.77[HA] = 250 mM
[HA] = 250/2.77 = 90.25 mM
[A-] = 1.77(90.25) = 159.74 mM
Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:

We know that,
The relation between the
for an ideal gas are :

As we are given :



Now we have to calculate the entropy change of the gas.


(b) As we know that, the work done for isochoric (constant volume) is equal to zero. 
(C) Heat during the process will be,

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Answer:
V = 48.5 L
Explanation:
Converting °C to K and kPa to atm
T = 25.0°C + 273.15 = 298.15 K
P = 61.3 kPa × (1 atm / 101.325 kPa) = 0.60498 atm
Calculating the volume of gas
V = nRT / P
V = (1.20 mol)(0.082057 L•atm/mol•K)(298.15 K) / 0.60498 atm
V = 48.5 L
Answer:
Pressure on Surface A > Pressure on Surface B
Explanation:
Pressu
re on Surface B > Pressure on Surface A