Question

Calculate the pH of 1.00 L of a solution that is 0.120M in HNO₂ and 0.150M in NaNO₂ before and after you add 2.0mL of 15.0M HCL. [K_a(HNO₂) = 4.0 x 10^-4]

Answer 1

its an acidic buffer as HNO2 is a weak acid and NaNO2 is a salt of weak acid HNO2 and strong base NaOH

moles of NaNO2 = 0.15 moles

moes of HNO2 = 0.12

pH = pKa + log [salt] / [acid]

pH = 3.4 + log [0.15/0.12 ]

pH = 3.4 + 0.097 = 3.5

now when you add HCl following reaction takes place ...

NaNO2 + HCl -----> HNO2 + NaCl

so HCl converts NaNO2 into HNO2 in other words it decreases concentration of NaNO2 and increases concentration of HNO2

now no. of moles of HCl added = 15.0 M x 2/1000 = 0.030 moles

since HCl and NaNO2 are reacting in 1:1 ratio

so 0.030 moles of HCl will convert 0.03 moles of NaNO2 into 0.030 moles of HNO2

so final no. of moles of HNO2 = 0.12 + 0.030 = 0.15 moles
final no. of moles of NaNO2 = 0.15 - 0.03 = 0.12

total volume = 1 L+ 0.002 L = 1.002 L

[salt] = [NaNO2] = [0.12 / 1.002L] = 0.1197 M
[acid] = [HNO2] = [0.15/1.002L] = 0.1497 M

pH = 3.4 + log [0.1197/0.1497]

pH =3.3