HCl + NaOH ---> NaCl + H20
If you follow guidance from other questions I have already answered for you, you will see that the above equation is balanced as it is.
P
H
=
−
log
10
[
H
3
O
+
]
=
−
log
10
{
2.3
×
10
−
6
}
=
−
{
−
5.64
}
=
5.64
Answer:
Mass percentage → 0.074 %
[F⁻] = 741 ppm
Explanation:
Aqueous solution of flouride → [F⁻] = 0.0390 M
It means that in 1L of solution, we have 0.0390 moles of F⁻
We need the mass of solution and the mass of 0.0390 moles of F⁻
Mass of solution can be determined by density:
1g/mL = Mass of solution / 1000 mL
Note: 1L = 1000mL
Mass of solution: 1000 g
Moles of F⁻ → 0.0390 moles . 19g /1 mol = 0.741 g
Mass percentage → (Mass of solute / Mass of solution) . 100
(0.741 g / 1000 g) . 100 = 0.074 %
Ppm = mass of solute . 10⁶ / mass of solution (mg/kg)
0.741 g . 1000 mg/1g = 741 mg
1000 g . 1 kg/1000 g = 1kg
741 mg/1kg = 741 ppm
As I am reading the problem, I see they gave you two pressures, one volume and they are asking for another volume. this should give you a hint that you need to use the following formula.
P1V1= P2V2
P1= 1.00 atm
V1= 0.50 ft³
P2= 3.00 atm
V2= ?
Now we plug the values
(1.00 x 0.50)= (3.00 x V2)
V2= 0.17 ft³
Use 25.4 g/ 425ml x 100 5.94 percent
