Initial mol HNO2
Volume of NaOH added
OH- + HNO2 ----- H2O + NO2-
Before addition 0 0.00375mol ----- 0
Addition 0.00375mol --- --- ---
After addition 0 0 0.00375mol
[NO2-]= 0.00375mol = 0.00375mol = 0.075M
Total volume added 0.025L + 0.025L
NO2- + H2O ----- HNO2 + OH-
I 0.075M --- 0 0
C -x ---- x x
E 0.075-x x x
ka= 4.50e-4
kb= kw/ka = 10-4/4.5e-4= 2.22e-11
kb= [HNO2][OH] = x2____
[NO2-] 0.075-x
2.22e-11= x2/0.075-x X= 1.29e-6 pOH= -log(1.29e-6) = 5.89
Solve for x, get rid of the x at the bottom, multiply 2.22e-11 by 0.075 and take the square root.
pH= 14-5.89
pH= 8.11