Answer:
The Ka is 9.11 *10^-8
Explanation:
<u>Step 1: </u>Data given
Moles of HX = 0.365
Volume of the solution = 835.0 mL = 0.835 L
pH of the solution = 3.70
<u>Step 2:</u> Calculate molarity of HX
Molarity HX = moles HX / volume solution
Molarity HX = 0.365 mol / 0.835 L
Molarity HX = 0.437 M
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<u>Step 3:</u> ICE-chart
[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4
Initial concentration of HX = 0.437 M
Initial concentration of X- and H3O+ = 0M
Since the mole ratio is 1:1; there will react x M
The concentration at the equilibrium is:
[HX] = (0.437 - x)M
[X-] = x M
[H3O+] = 1.995*10^-4 M
Since 0+x = 1.995*10^-4 ⇒ x=1.995*10^-4
[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M
[X-] = x = 1.995*10^-4 M
<u>Step 4: </u>Calculate Ka
Ka = [X-]*[H3O+] / [HX]
Ka = ((1.995*10^-4)²)/ 0.437
Ka = 9.11 *10^-8
The Ka is 9.11 *10^-8
I believe the correct answer from the choices listed above is option A. A direct relationship can be represented by the expression:
y/x = k
<span>A direct relationship is an expression between two numbers or other variables where an increase or decrease in one variable causes the same change in the other value.</span>
The pure form of sodium hydroxide is solid. Compared to the pure form, the aqueous form of the liquid is actually many times more volatile. For this very reason, it is advised that the bottles used to keep samples of NaOH be kept closed.
From ideal gas equation that is PV=nRT
n(number of moles)=PV/RT
P=760 torr
V=4.50L
R(gas constant =62.363667torr/l/mol
T=273 +273=298k
n is therefore (760torr x4.50L) /62.36367 torr/L/mol x298k =0.184moles
the molar mass of NO2 is 46 therefore density= 0.184 x 46=8.464g/l
