Question

What is the pH of a solution prepared by adding 1.12g of HNO₂ to 315.00 mL of water? (Acid Ionization Constant for HNO₂ = 4.50E-4)

a) 3.73
b) 11.77
c) 7.00
d) 2.23
e) 5.83E-3

Answer 1

mole 47.0 g Concentration of HNO, is => L 315.00 mL 1000 mL =0.0757M Kaof HNO, is 4.50 x104 Hydrolyis of the acrylic acid reaction is, HNO, + 1,0 = NO, +H,0* Initial 0.0757 0 0 Change Equilbrium 0.0757 - x Now, substitute these concentrations and the value of Ka into the equilibrium-constant equation: [NO, H20] Ka=! [HNO2] -X Xxx 4.5 x 104 - 0.0757 - x Assume x is much smaller than 0.0757, so (0.0757 – x)= 0.0757 x2 = 4.5x10-4 x0.0757 = 34.0x10 x=(34.0x10-6 )] =5.83x10-3 x=[H0*]=5.83x10-3 pH=- log[H,0*] =- log(5.83x10-) = 3 – log(5.83) = 2.23 Therefore, pH of HNO, is 2.23