We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+ + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:
PH= -㏒[H+]
= -㏒(4.81x10^-7) = 6.32
Answer:
The concentration of this sodiumhydroxide solutions is 0.50 M
Explanation:
Step 1: Data given
Mass of sodium hydroxide (NaOh) = 8.0 grams
Molar mass of sodium hydroxide = 40.0 g/mol
Volume water = 400 mL = 0.400 L
Step 2: Calculate moles NaOH
Moles NaOH = mass NaOH / molar mass NaOH
Moles NaOH = 8.0 grams / 40.0 g/mol
Moles NaOh = 0.20 moles
Step 3: Calculate concentration of the solution
Concentration solution = moles NaOH / volume water
Concentration solution = 0.20 moles / 0.400 L
Concentration solution = 0.50 M
The concentration of this sodiumhydroxide solutions is 0.50 M
Answer:
Volume of water at this temperature is 27.2 mL
Explanation:
We know that 
Here density of water is 0.992 g/mL
Here mass of water is 27.0 g
So 
= 
= 27.2 mL
Answer:
15.04 mL
Explanation:
Using Ideal gas equation for same mole of gas as
Given ,
V₁ = 21 L
V₂ = ?
P₁ = 9 atm
P₂ = 15 atm
T₁ = 253 K
T₂ = 302 K
Using above equation as:
Solving for V₂ , we get:
<u>V₂ = 15.04 mL</u>
