Question

Calculate the pH of the buffer solution (1.00 M HNO2 and 1.00 M NaNO2) after 0.20 mol of NaOH is added to 1.00 L of buffer solution (Ka = 4.0 x 10^-4)

Answer 1

when NaOH is added, it will react with HNO2 to convert to NaNO2. So, amount of HNO2 that has reacted with naOH has to be substracted from HNO2 and has to be added to NaNO2.

moles of HNO2 present initially = 1M *1 L = 1mol

moles of NaNo2 present initally = 1mol

moles of base added = moles of naNO2 formed = 0.2 mol

moles of HNO2 present after adding NaOH = 1-0.2 = 0.8

moles of NaNO2 present in the solution after adding NaOH = 1+0.2 = 1.2 moles

pka = -log ka = -log(4.0*10^-4) = 3.39

pH = pka +log[naNO2/HNO2]

= 3.39 + log[0.8/1.2]=3.21

pH of the solution after adding naOh will be 3.21.