In: Chemistry
I cannot figure out this problem, I've made a few attempts at solving it. I could really use some help. The question is : 75.00 mL of 0.225 M HNO2 is titrated to its equivalence point with 0.100 M NaOH. What is the pH at its' equivalence point? Please help!
HNO2 + NaOH ---------------------> NaNO2 + H2O
75x 0.225 75x0.1 0 0
16.875 7.5 0 0 ------------------initial millimoles
9.375 0 7.5 7.5 -------------- at equivalent point
in the solution HNO2 + NaNO2 remained . so it is buffer
For acidic buffer
Henderson-Hasselbalch equation
pKa = -log Ka = -log(4.0 x 10-4) = 3.39
pH = pKa + log[salt/acid]
pH = 3.39 + log[7.5/9.375]
pH = 3.29