First, we need to get the concentration of [NaH2PO4]:
[NaH2PO4] =( mass / molar mass ) * volume L
when we have mass NaH2PO4 = 6.6 g & molar mass = 120g/mol & V = 0.355 L
So by substitution:
[NaH2PO4] = (6.6g / 120g/mol) * 0.355 L = 0.0195 M
then, we need to get the concentration of [Na2HPO4]:
[Na2HPO4]= (mass / molar mass ) * volume L
So by substitution:
[Na2HPO4] = (8g/ 142g/mol) * 0.355 L = 0.02 M
and when Pka of the 2nd ionization of phosphoric acid = 7.21
So by substitution in the following formula, we can get the PH:
PH = Pka + ㏒[A]/[AH]
∴PH = 7.21 + ㏒[0.02]/[0.0195]
∴ PH = 7.2
Answer:
The volume of STP will be 0.03 plss mark me brainliest
Answer : The pressure of the gas using both the ideal gas law and the van der Waals equation is, 60.2 atm and 44.6 atm respectively.
Explanation :
First we have to calculate the pressure of gas by using ideal gas equation.

where,
P = Pressure of
gas = ?
V = Volume of
gas = 0.805 L
n = number of moles
= 1.93 mole
R = Gas constant = 
T = Temperature of
gas = 306 K
Now put all the given values in above equation, we get:


Now we have to calculate the pressure of gas by using van der Waals equation.

P = Pressure of
gas = ?
V = Volume of
gas = 0.805 L
n = number of moles
= 1.93 mole
R = Gas constant = 
T = Temperature of
gas = 306 K
a = pressure constant = 
b = volume constant = 
Now put all the given values in above equation, we get:
![(P+\frac{(4.19L^2atm/mol^2)\times (1.93mole)^2}{(0.805L)^2})[0.805L-(1.93mole)\times (5.11\times 10^{-2}L/mol)]=1.93mole\times (0.0821L.atm/mol.K)\times 306K](https://tex.z-dn.net/?f=%28P%2B%5Cfrac%7B%284.19L%5E2atm%2Fmol%5E2%29%5Ctimes%20%281.93mole%29%5E2%7D%7B%280.805L%29%5E2%7D%29%5B0.805L-%281.93mole%29%5Ctimes%20%285.11%5Ctimes%2010%5E%7B-2%7DL%2Fmol%29%5D%3D1.93mole%5Ctimes%20%280.0821L.atm%2Fmol.K%29%5Ctimes%20306K)

Therefore, the pressure of the gas using both the ideal gas law and the van der Waals equation is, 60.2 atm and 44.6 atm respectively.