Data:
Molar Mass of HNO2
H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of HNO2 = 1 + 14 + 48 = 63 g/mol
M (molarity) = 0.010 M (Mol/L)
Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) = [tex]5.0*10^{-4} [/tex]
[tex] \alpha^2 (degree\:of\:ionization) = ?[/tex]
[tex]Ka = M * \alpha^2[/tex]
[tex]5.0*10^{-4} = 0.010* \alpha^2 [/tex]
[tex]0.010\alpha^2 = 5.0*10^{-4}[/tex]
[tex]\alpha^2 = \frac{5.0*10^{-4}}{0.010} [/tex]
[tex]\alpha^2\approx500*10^{-4}[/tex]
[tex]\alpha\approx\sqrt{500*10^{-4}} [/tex]
[tex]\alpha \approx 2.23*10^{-3}[/tex]
Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:
[tex][ H_{3} O^+] = M* \alpha [/tex]
[tex][ H_{3} O^+] = 0.010* 2.23*10^{-3}[/tex]
[tex][ H_{3} O^+] \approx 0.0223*10^{-3}[/tex]
[tex][ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L[/tex]
And finally, we will use the data found and put in the logarithmic equation of the PH, thus:
Data:
log10(2.23) ≈ 0.34
pH = ?
[tex][ H_{3} O^+] = 2.23*10^{-5}[/tex]
Formula:
[tex]pH = - log[H_{3} O^+][/tex]
Solving:
[tex]pH = - log[H_{3} O^+][/tex]
[tex]pH = -log2.23*10^{-5}[/tex]
[tex]pH = 5 - log2.23[/tex]
[tex]pH = 5 - 0.34[/tex]
[tex]\boxed{\boxed{pH = 4.66}}\end{array}}\qquad\quad\checkmark[/tex]
Note:. The pH <7, then we have an acidic solution.
