Data:
Molar Mass of HNO2
H = 11 = 1 amu
N = 114 = 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) = \(5.0*10^{-4}\)
\(\alpha^2 (degree\:of\:ionization) = ?\)
\(Ka = M * \alpha^2\)
\(5.0*10^{-4} = 0.010* \alpha^2\)
\(0.010\alpha^2 = 5.0*10^{-4}\)
\(\alpha^2 = \frac{5.0*10^{-4}}{0.010}\)
\(\alpha^2\approx500*10^{-4}\)
\(\alpha\approx\sqrt{500*10^{-4}}\)
\(\alpha \approx 2.23*10^{-3}\)
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:
\([ H_{3} O^+] = M* \alpha\)
\([ H_{3} O^+] = 0.010* 2.23*10^{-3}\)
\([ H_{3} O^+] \approx 0.0223*10^{-3}\)
\([ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L\)
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 = ?
\([ H_{3} O^+] = 2.23*10^{-5}\)
Formula:
\(pH = - log[H_{3} O^+]\)
Solving:
\(pH = - log[H_{3} O^+]\)
\(pH = -log2.23*10^{-5}\)
\(pH = 5 - log2.23\)
\(pH = 5 - 0.34\)
\(\boxed{\boxed{pH = 4.66}}\end{array}}\qquad\quad\checkmark\)
Note:. The pH <7, then we have an acidic solution.
